UDENTS 
CULTURE 

SOR  •  ^ 


MATHEMATICS 
FOR  STUDENTS  OF  AGRICULTURE 


A  SERIES  OF  MATHEMATICAL  TEXTS 

EDITED   BY 

EARLE   RAYMOND  HEDRICK 


THE  CALCULUS 

By    Ellbry    Williams     Davis     and     William    Charles 
Brenke. 

ANALYTIC  GEOMETRY  AND  ALGEBRA 

By  Alexander  Ziwet  and  Louis  Allen  Hopkins. 

ELEMENTS  OF  ANALYTIC  GEOMETRY 

By  Alexander  Ziwet  and  Louis  Allen  Hopkins. 

PLANE  AND  SPHERICAL  TRIGONOMETRY 

By  Alfred  Monroe  Kenyon  and  Louis  Ingold. 

ELEMENTS  OF  PLANE  TRIGONOMETRY 

By  Alfred  Monroe  Kenyon  and  Louis  Ingold. 

ELEMENTARY  MATHEMATICAL  ANALYSIS 

By  John  Wesley  Young  and  Frank  Millett  Morgan. 

PLANE  TRIGONOMETRY 

By  John  Wesley  Young  and  Frank  Millett  Morgan. 

COLLEGE  ALGEBRA 

By  Ernest  Brown  Skinner. 

MATHEMATICS    FOR    STUDENTS    OF    AGRICULTURE   AND 
GENERAL  SCIENCE 
By    Alfred    Monroe     Kenyon     and    William    Vernon 

LOVITT. 

MATHEMATICS  FOR  STUDENTS  OF  AGRICULTURE 
By  Samuel  Eugene  Rasor. 

THE  MACMILLAN  TABLES 

Prepared  under  the  direction  of  Earle  Raymond  Hedrick. 

PLANE  AND  SOLID  GEOMETRY 

By  Walter  Burton  Ford  and  Charles  Ammerman. 

CONSTRUCTIVE  GEOMETRY 

Prepared  under  the  direction  of  Earle  Raymond  Hedrick. 

JUNIOR  HIGH  SCHOOL  MATHEMATICS 

By  William  Ledley  Vosburgh  and  Frederick  William 
Gentleman. 


MATHEMATICS 


FOR 


STUDENTS  OF  AGRICULTURE 


BY 

SAMUEL   EUGENE   RASOR 

PROFESSOR    OF    MATHEMATICS    IN 
THE    OHIO    STATE    UNIVERSITY 


NrtD  gotft 

THE  MACMILLAN   COMPANY 
1931 


POINTED  IN  THE   UNITED   STATES   OF  AMERICA 


Copyright,  1921, 

Bt  the  macmillan  company. 


All  rights  reserved  —  no  part  of  this 

book  may  be  reproduced  in  any  form 

without  permission  in  writing  from 

the  publisher. 


Set  up  and  electrotyped.    Published  September,  1921. 


NorbjDoti  Press 

J.  S.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


PREFACE 

This  book  presents  a  year's  work  in  mathematics  for  students 
taking  agricultural  courses  in  secondary,  vocational,  and  technical 
schools  and  in  colleges  and  universities.  By  omitting  the  more  ele- 
mentary or  the  advanced  topics,  as  may  seem  desirable,  or  by  prop- 
erly selecting  the  exercises  to  be  solved,  the  book  will  serve  quite 
naturally  for  a  half-year's  work. 

The  modem  agriculturalist  must  devote  a  larger  and  larger  part 
of  his  time  to  mental  activity  along  scientific  lines.  To  do  this 
rationally  he  must  have  better  methods  for,  and  more  practice  in, 
estimating,  computing,  and  comparing  the  materials  and  processes 
with  which  he  works.  The  object  of  the  present  text  is  to  furnish 
a  better  basis  for  effecting  these  ends. 

The  text  has  been  used  in  the  form  of  mimeographed  notes  for 
several  years  in  the  classroom,  and  has  been  modified  from  time  to 
time  in  the  Hght  of  experience  and  of  new  conditions.  It  was  soon 
found  that  it  was  not  safe  to  assimie  too  much  preparation  on  the 
part  of  students  taking  the  course.  Consequently,  material  is  pre- 
sented in  the  text  for  study  and  practice  in  basic  elementary  pro- 
cesses in  arithmetic,  algebra,  and  geometry.  Some  study  of  this 
material  will  suffice  for  those  students  who  have  not  had  the  proper 
preparation  for  the  course  as  well  as  for  those  students  who  wish  to 
review  elementary  facts  and  principles. 

The  unifying  element  in  the  book  is  the  attempt  to  make  the  prin- 
ciples of  arithmetic,  algebra,  geometry,  trigonometry,  and  graphic 
representation  fimction  with  the  student's  interest  and  point  of  view. 
It  is  hoped  that  in  this  way  he  may  find  a  natural  tendency  to  use 
this  most  basic  science  in  the  solution  at  least  of  those  problems 
which  arise  in  his  own  sphere  of  activity.  We  may,  therefore,  expect 
the  student  to  acquire  in  this  way  not  less  of  mathematical  value 
and  content  but  more  of  it. 

The  following  are  features  of  various  chapters: 

Chapter  I  presents  suggestions  in  drawing  as  a  means  of  review- 
ing facts  from  geometry  and  connecting  them  naturally  with  such 
observations  as  the  student  can  make;  for  example,  estimating  and 
computing  the  grade  of  a  roof  or  a  hill  or  a  ditch  or  a  railroad,  the 


910132 


VI 


PREFACE 


projections  of  the  sides  of  a  farm  contour  on  a  base  line,  scale  draw- 
ings of  farms  and  farm  buildings,  and  the  composition  and  resolu- 
tion of  forces  graphically. 

In  Chapter  II  a  basis  is  presented  for  rational  calculation  and  for 
estimation,  including  the  question  of  accuracy.  The  topics  consid- 
ered are  percentage,  profit  and  loss,  bank  discount,  elementary  men- 
suration of  common  plane  figures  and  solids,  including  rules  for  esti- 
mating silos. 

In  Chapter  III  a  treatment  of  indirect  measurement  and  propor- 
tion is  given.  This  includes  also  the  study  of  the  sine,  cosine,  and 
tangent  ratios  and  tables  for  them,  with  applications  to  simple  prob- 
lems in  farm  surveying. 

A  review  of  the  main  facts  of  elementary  algebra  may  be  obtained 
from  Chapter  IV.  Graphic  representation  of  quantities,  including 
price  curves,  weight  and  age  curves,  leveling,  etc.,  is  presented  in 
Chapter  V.  A  rather  fundamental  treatment  of  compound  interest, 
annuities,  depreciation,  and  farm  loans  is  given  in  Chapter  IX. 

Chapter  X  treats  averages  and  mixtures.  It  includes  a  discussion 
of  dairy  problems,  arithmetic  and  geometric  means,  weighted  aver- 
ages, the  median,  the  mode,  fertilizer  mixtures,  and  cement  mixtures. 

The  chapter  on  Land  Surveying  (Chapter  XIII)  may  be  of  inter- 
est from  the  standpoint  of  the  applications  and  the  new  material. 

Chapter  XIV  is  entitled  Simple  Machines.  It  presents  a  treat- 
ment of  levers,  the  wheel  and  axle,  the  inclined  plane,  the  screw, 
pulleys,  safe  loads  for  beams,  work,  horse-power,  working  day  for  a 
horse,  draft,  etc. 

The  tables  printed  in  the  Appendix  at  the  end  of  the  book  are 
adapted  to  the  requirements  of  computing  and  estimating  the  exer- 
cises and  examples.  Throughout  the  text  the  exercises  and  illustra- 
tions are  drawn  largely  from  experience  and  the  point  of  view  of  the 
agriculturahst. 

The  author  desires  to  acknowledge  his  indebtedness  to  Dean  Al- 
fred Vivian  and  members  of  the  faculty  of  the  College  of  Agriculture 
of  The  Ohio  State  University,  to  R.  D.  Bohannan,  Professor  of 
Mathematics  in  The  Ohio  State  University,  and  to  E.  R.  Hedrick, 
Professor  of  Mathematics  in  the  University  of  Missouri,  for  the  in- 
spiration of  their  interest  and  suggestions. 

S.  E.  Rasor. 

The  Ohio  State  University. 


CONTENTS 


PAGE 


Chapter  I.    Drawing  —  Graphic  Solutions 1 

Chapter  II,     Computation  —  Measurement 10 

Chapter  III.    Indirect  Measurement — Trigonometry — Survey- 
ing         44 

Chapter  IV.    Review  of  Algebra 63 

Chapter  V.     Graphic  Representation  of  Quantities       ...  88 

Chapter  VI.     Graphs  in  Algebra 104 

Chapter  VII.     Computation  by  Logarithms 124 

Chapter  VIII.     The  Progressions 139 

Chapter  IX.    Compound  Interest  —  Annuities  —  Depreciation    .  149 

Chapter  X.     Averages  and  Mixtures 167 

Chapter  XI.    Geometry  —  Mensuration       .        .        .        .        .  184 

Chapter  XII.     Oblique  Triangles 200 

Chapter  XIII.     Land  Surveying 218 

Chapter  XIV.     Simple  Machines 235 

Chapter  XV.    Composition  and  Resolution  of  Forces  .        .        .  261 


vu 


A  LIST  OF  SIGNS  AND  SYMBOLS 

+  ,  redid  plus.         — ,  read  minus. 

X ,  or  •  ,  read  times. 

-T- ,  read  divided  by. 

=,  read  equals,  or  is  equal  to. 

z^,  read  is  not  equal  to. 

<,  read  is  less  than. 

>,  read  is  greater  than. 

.'.,  read  therefore. 

',  redid  feet,  e.g.  2',  means  2  feet. 

",  read  inches,  e.g.  2",  means  2  inches. 

^  ^'  '  These  are  signs  of  aggregation.     The  expressions 

which  they  inclose  are  to  be  treated  together  as  one 


quantity. 


[  ],  Brackets. 

{  },  Braces. 

a  :  b,  read  a  is  to  b. 

Va,  read  square  root  of  a. 

Va,  read  n'*  root  of  a. 

a^,  read  a  to  the  n^*  power,  or  a  exponent  n. 

log  oW,  read  logarithm  of  n  to  the  base  a. 

±'r,  read  perpendicular,  or  is  perpendicular  to 

Z,  read  angle,   A,  read  angles. 

A,  read  triangle,  A,  read  triangles. 

Kt.  A,  read  right  triangle. 

O,  redid  parallelogram. 


MATHEMATICS  FOR  STUDENTS 
OF  AGRICULTURE 

CHAPTER  I 

INTRODUCTION  —  DRAWING  —  GRAPHIC 
SOLUTIONS 

1.  Instruments  and  Materials.  A  large  number  of  the 
exercises  and  problems  in  this  book  are  to  be  solved  graph- 
ically, i.e.  by  drawing  appropriate  diagrams.  To  do  this 
work  neatly  and  accurately,  a  few  simple  drawing  instru- 
ments are  essential.  The  materials  and  instruments  needed 
for  the  work  are  described  briefly  below.* 

(a)  Pencils.  One  medium  hard  pencil  sharpened  to  a 
round  point  may  be  used  for  lettering.  A  hard  pencil 
sharpened  to  a  round  point  may  be  used  for  sketch- 
ing and  marking  points  on  paper.  Another  hard 
pencil,  sharpened  to  a  chisel-point  (Fig.  1),  may  be 
used  to  draw  straight  lines.  A  piece  of  sandpaper 
may  be  used  to  whet  the  points. 

(6)  A  drawing  hoard,  about  12  or  15  by  18  inches. 
It  should  be  made  of  soft  wood  and  have  at  least 

.   ,  ,  Fig.  1 

one  straight  edge. 

(c)  A  T-square,  15  to  18  inches  long.  As  shown  in 
Fig.  2,  a  T-square  consists  simply  of  a  wooden  strip,  Ay 

♦Those  students  who  are  taking  courses  in  drawing  are  already  in  posses- 
sion of  many  of  these  instruments  and  materials  needed  for  graphical  work. 

1 


MATHEMATICS 


[I,  §1 


.A     1 


called:  th>bhead'f  to.  wfeich'  is  fastened  a  thin,  straight-edged 
blade,  B.  '  '         ' . 

(d)  Two  trari'spa'irent  Mangles.  Each  of  these  should  have 
one  right  angle.  The  acute  angles  of  one  triangle  should 
be  45°,  while  the  acute  angles  of  the  r 
other  should  be  30°  and  60°  (Fig.  3).  ^ 

{e)  A  scale  with  the  unit  of  length 
divided  into  tenths  is  to  be  preferred  to 
one  with  the  unit  of  length  divided  into 
eighths,  twelfths,  or  sixteenths. 

(/)  A  "protractor,  Fig.  3,  4  to  6  inches 
in  diameter  is  recommended.  It  consists 
simply  of  angular  graduations  upon  a 
semicircular  piece  of  metal  or  some  trans- 
-parent  material.  It  is  used  for  laying  off 
and  measuring  angles  upon  a  drawing. 

ig)  Two  pairs  of  compasses.  One  should 
be  a  pair  of  6-inch  pencil  compasses, 
the  other  should  be  inking  compasses 
of  the  same  size. 

Qi)    A   pair  of   6-inch  hairspring  di- 
viders.   These  are  to  be  used  in  calculating  lengths,  distances, 
etc.  on  drawings  made  to  scale. 


T-SQUARE 

Fig.  2 


30°  Triangle 


45°  Triangle 


Protractor 


Fig.  3 


I,  §  2]  INTRODUCTION  3 

2.  General  Directions.  All  work  must  be  done  with 
neatness  and  accuracy.  The  acquisition  of  these  two 
virtues  in  graphical  work  may  require  an  unusual  amount 
of  time  and  effort.  However,  when  they  are  once  estab- 
lished as  habits,  much  time  will  be  saved  in  avoiding  and 
in  detecting  errors,  as  well  as  in  checking  results. 

ABCDErGH/JKLMNOF^OnSTUVWXYZ 
obc  defgh  ijk/mnop  qrstuv  wxyz 

/234567a90&. 

Fig.  4.  —  A  Set  of  Letters  and  Figures 

All  drawings  should  be  done  in  pencil.  After  some  train- 
ing in  the  use  of  the  ruling  pen,  the  drawings,  if  it  is  so  desired, 
may  be  "  inked  in  "  with  drawing  ink. 

Figure  4  shows  a  set  of  letters  and  figures.  They  may  be 
used  in  lettering  descriptive  work  on  a  drawing  instead  of 
using  ordinary  writing. 

EXERCISES 

1.  Draw  a  straight  line.     For  this  purpose  two  things  are  required : 
(a)  A  plane  or  flat  surface  on  which  to  draw  the  line. 

(6)  A  straightedge  to  be  used  as  a  guide. 

2.  State  some  facts  about  a  straight  line  by  testing  a  straightedge. 
To  do  this,  select  two  points  through  which  the  straightedge  is  to  pass, 
draw  a  line  through  them,  reverse  the  straightedge,  and  redraw  the 
line. 

How  many  points  determine  a  straight  line?  How  many  points 
may  two  distinct  straight  lines  have  in  common  ?  When  does  a  straight 
line  lie  wholly  in  a  plane? 

3.  On  what  surfaces,  other  than  the  plane,  is  it  possible  to  lay  a 
straight  line  so  that  each  of  its  points  coincides  with  the  surface  ? 

In  this  connection,  examine  a  cylinder ;  a  cone ;  a  riding  saddle ;  a 
capstan. 

4.  The  part  of  a  straight  line  between  two  of  its  points  is  called  a 
line  segment  or  simply  a  segment,  and  is  designated  by  its  end  points, 


4  MATHEMATICS  [I,  §  2 

or  sometimes  by  a  small  letter  near  the  middle  of  the  segment.    Thus  we 
speak  of  the  ''segment  AB''  or  the  "segment  a."     See  Fig.  5. 


Fig.  5 

On  a  given  straight  line  construct  a  segment  equal  to  a  given  seg- 
ment.    Do  this  in  two  ways. 

(a)  By  the  use  of  a  pair  of  dividers. 

(b)  By  the  use  of  a  scale  or  a  straightedge,  —  perhaps  an  inch 
scale,  or  a  centimeter  scale. 

Which  of  these  methods  is  the  more  accurate?  How  could  the 
accuracy  of  the  second  method  be  improved? 

5.  Locate  any  four  points  A,  B,  C,  D  in  the  plane  and  draw  the 
straight  lines  which  any  pair  of  them  determines.  In  general,  how 
many  of  these  lines  are  there?  Under  what  conditions  are  there  less 
than  this  number  of  lines  ? 

6.  Draw  four  straight  lines  at  random.  In  general  how  many 
points  of  intersection  are  there?  When  are  there  less  than  this  num- 
ber of  intersections?  What  are  the  names  of  some  of  the  possible 
figures  obtained  in  this  way? 

7.  Given  two  unequal  segments  a  and  b.  Draw  the  segment  equal 
to  a-\-b.  Equal  to  a  —  b  when  a>b.  Draw  the  segment  equal  to  a—b 
when  a<b. 

8.  Draw  the  shortest  distance  between  two  given  points  A  and  B 
in  a  plane. 

9.  Determine  all  the  points  which  are  the  same  distance  from 
two  given  points.  For  example,  how  could  a  fence  be  located  so 
that  any  place  on  it  would  be  equidistant  from  a  certain  house  and 
barn? 

10.  Determine  the  point  C  midway  between  two  points  A  and  B. 
Draw  a  perpendicular  to  C  by  using  dividers  and  ruler.  Find  the 
mid-point  C  by  using  the  draftsman's  triangle  alone. 

11.  Draw  the  shortest  distance  from  a  point  A  to  a  given  straight 
line.     From  a  point  A  to  a  given  circle. 

12.  By  use  of  ruler  and  compass,  bisect  a  given  angle.  Draw  angles 
of  90°,  45°,  135°,  22|°. 

13.  Solve  the  previous  exercise  by  use  of  a  carpenter's  square  alone. 

14.  Draw  in  a  circle,  by  use  of  ruler  and  compass,  an  inscribed  square. 
An  inscribed  octagon.     An  inscribed  hexagon. 


I,  §3] 


INTRODUCTION 


M- 


A 


B^^jC 


N 


FiQ.  6 


15.  Draw,  with  ruler  and  compass,  right  triangles  with  acute  angles 
30°,  45°,  60°. 

16.  How  are  the  sides  related  in  each  of  the  triangles  in  the  pre- 
ceding exercise? 

Do  these  relations  in  any  way  describe  or  determine  the  angles  of 
these  triangles,  i.e.  if  these  ratios  of  the  sides  in  a  right  triangle 
were  given,  would  you  then  be  able  to  discover  , 

what  angles  they  determined  ?  ^-;^ 

17.  Prove  that  the  image  of  an  object  as  seen  '     ^C^^^ 
in  a  plane  reflecting  surface  appears  to  be  as  far 
back  of  the  surface  as  the  object  is  in  front  of  it, 
using  the  law  of  reflection  that  the  angle  of  in- 
cidence is  equal  to  the  angle  of  reflection. 

Note.  The  angle  of  incidence  is  the  angle 
which  the  path  of  a  ray  of  light  makes  with  the  perpendicular  to  the 
reflecting  surface  as  it  strikes  the  surface.  The  angle  of  reflection  is 
the  angle  the  reflected  ray  makes  with  the  per- 
pendicular as  it  leaves  the  surface.  In  the 
accompanying  figure  i  =  the  angle  of  incidence 
and  r  =  the  angle  of  reflection.  Light,  sound, 
and  moving  particles  are  reflected  according  to 
the  law  that  Zi  =  Zr. 

18.  Determine  the  path  of  a  ray  of  light  whose  origin  is  at  a  given 
point  A  and  which  is  reflected  from  a  given  straight  line  MN  to  a  given 
point  B. 

19.  A  ladder  leaning  against  a  barn 
slips  to  the  ground.  In  doing  so, 
what  curve  does  the  center  of  the 
ladder  describe  if  the  foot  of  the  lad- 
der moves  directly  and  horizontally  away  from  the  barn? 

[Hint.  How  far  from  the  vertex  of  the  right  angle  is  the  center  of 
the  hypotenuse  in  any  right  triangle?] 


Fig.  7 


M. 


.N 


Fig.  8 


3.  Drawing  to  Scale.  Since  it  is  not  possible  in  most 
cases  to  draw  objects  full  size  on  paper,  the  drawings  are 
made  proportionally  smaller.  This  consists  merely  in  making 
all  the  dimensions  of  the  drawing  a  certain  fraction  of  the 
true  dimensions  of  the  object.    This  is  called  drawing  to 


6  MATHEMATICS  [I,  §  3 

scale.  An  example  is  the  plan  of  a  house  drawn  to  the  scale  : 
1  inch  (on  the  drawing)  =  2  feet  (on  the  house),  i,e.  to  the 
scale  of  1  in  24  or  -^. 

A  large  number  of  problems  may  be  solved  graphically  by 
drawing  the  given  data  to  scale  and  then  measuring  the 
unknown  parts.  Rather  accurate  results  are  readily  obtained 
in  this  way.  It  also  furnishes  a  desirable,  independent 
method  of  checking  the  calculations  in  problems  for  which 
an  accurate  solution  has  been  obtained  otherwise. 

EXERCISES 

1.  Draw  a  ground  plan  of  your  home  farm  lot  showing  houses, 
barn,  etc. ;  or  such  a  plan  of  your  city  or  suburban  home. 

2.  Draw  a  ground  plan,  scale  1"  =  10',  of  a  barn  40' X50'  to  house 
4  horses  and  10  cows,  and  with  arrangements  in  it  for  tools  and  harness 
and  extra  pens  for  calves,  etc.* 

3.  Three  miles  from  a  straight  road  is  a  fort  with  a  gun  whose  range 
is  five  miles.     Find,  by  drawing  to  the  scale  of  \"  to  2  miles,  what 

distance   along  this   road   is   com-      ^ 

manded  by  the  gun.  5  x<^^A_.. tg//  ^ 

4.  Find,  by  a  scale  drawing,  the  5  y<<r  y'''^\^^^  \^  3^ 
height  of  a  tree  which,  at  a  distance         ><^\/^            \a^ 

of  350'  from  an  observer,  subtends        ^^ 

an  angle  of  22°.  ^^°-  ^ 

5.  How  can  a  carpenter  divide  a  4f  in.  board  into  3  equal  strips 
without  the  use  of  fractions?     See  Fig.  9. 

6.  Divide  a  given  line  segment  into  seven  equal  parts. 

4.  Projection  of  a  Line  Segment.  To  project  a  given 
segment,  as  AB,  Fig.  10,  upon  a  line  PQ  drop  perpendiculars 
to  PQ  from  the  end  points  of  the  segment.  Then  CD  is 
the  projection  "of  AB  upon  PQ  and  is  called  the  horizontal 

*  For  suggestions  on  the  architecture  of  farm  buildings,  see  L.  F.  Allen, 
Rural  Architecture,  published  by  L.  F.  Allen,  New  York;  Ekblaw,  Farm 
Structures,  Macmillan,  1916 ;  G.  G.  Hill,  Practical  Suggestions  for  Farm 
Buildings,  Wash.,  Gov.  Ptg.  Office,  1901 ;  I.  P.  Roberts,  The  Farmstead, 
Macmillan,  1900. 


I,  §4] 


INTRODUCTION 


projection  in  case  PQ  is  horizontal.  Projection  on  a  vertical 
line  is  called  vertical  projection  as  BE,  Fig.  10,  in  which  AE 
is  parallel  to  PQ,  and  BD  perpendicular  to  it. 


c  D       ^ 

FiQ.  10 

Similarly,  PQ  may  be  projected  upon  AB  (produced  if 
necessary) . 

In  the  case  of  a  hill,  or  a  grade,  or  a  rafter,  the  horizontal 
projection  is  called  the  run  of  the  rafter,  etc.,  and  the  vertical 
projection  is  called  the  rise.  The  rise  divided  by  run  is 
called  the  slope;  it  is  a  measure  of  the  steepness  of  the  hill, 
or  grade,  or  rafter.     For  example,  if  a  rafter  on  a  building 

Ridge 


„  Rise 

Slope  =^^ 

Run 


24'  wide  rises  6'  at  the  comb  of  the  roof,  we  say  it  rises  6'  in 
12'  or  1  in  2,  i.e.  the  slope  is  ^.  If  a  hill  rises  30'  in  a  run  of 
100',  it  has  a  slope  of  30  in  100,  i.e.  the  grade  is  30/100  or 
30%.  A  1%  grade,  as  for  a  steam  railroad,  rises  1'  in  100'. 
With  reference  to  a  roof,  the  word  pitch  is  widely  used ; 
it  denotes  rise  of  the  roof  divided  by  the  span  or  whole 
width  of  the  building  (Fig.  11).  Notice  that  the  pitch  of  a 
roof  is  ha,lf  gf  the  slope. 


8 


MATHEMATICS 


[I,  §4 


EXERCISES 
Draw  each  of  the  following  figures  on  an  enlarged  scale  and  project 
each  segment  upon  the  other.     Specify  the  projection  in  each  case. 
Find  the  horizontal  and  the  vertical  projections,  and  find,  by  measuring, 
the  per  cent  grade  for  each  segment. 

c 


3 


Fig.  12 


6.  The  altitude  of  an  equilateral  triangle  whose  side  is  12'  is  pro- 
jected upon  one  of  the  adjacent  sides.     Find  the  projection. 

7.  Find,  by  drawing  to  scale,  the  side  of  a  square  whose  diagonal 
is  10  rd. 

8.  Show  how  to  bisect  a  given  angle  by  using  two  carpenter's  squares. 
[Hint.     Use  the  geometric  theorem  that  any  point  on  the  bisector 

of  an  angle  is  equidistant  from  the  sides.] 

9.  Draw  any  oblique  triangle.     Draw  the  projections  of  two  sides 
on  the  third  side. 

5.  Graphic  Resolution  of  Forces.  A  force  may  be  repre- 
sented graphically  by  a  straight  line  whose  length  represents 
the  magnitude  of  the  force,  and  whose 
direction  is  that  of  the  force.     See  §  213. 

In  Fig.  13,  F  represents  a  force  acting 
on  W  in  the  direction  shown  and  at  an 
angle  BAC  with  the  horizontal. 

Project  F  on  a  horizontal  line  through  the  point  where  the 
force  is  applied  to  W.    In  this  way  the  horizontal  projection 


Fig.  13 


I,  §  5]  INTRODUCTION  9 

AB  is  said  to  be  the  horizontal  component  of  the  force  F 
and  the  perpendicular  BC  is  called  the  vertical  component 
of  the  force.  F  is  thus  resolved  into  vertical  and  horizontal 
components. 

EXERCISES 

1.  A  force  of  10  pounds  acts  on  a  weight  W  at  an  angle  of  30°  with 
the  horizontal.  Draw  the  figure,  making  F  10  units  in  length.  Project 
F  on  the  horizontal  line  through  the  point  of  application  of  the  force. 

By  the  scale,  determine  the  magnitude  of  each  component  in  pounds. 
What  is  thus  the  amount  and  the  direction  of  the  horizontal  and  the 
vertical  pull  on  TF  ? 

2.  In  drawing  a  sled  a  horse  pulls  200  pounds  on  his  traces,  which 
make  an  angle  of  10°  with  the  horizontal.  Plot  to  scale  the  figure 
representing  these  conditions  and  determine  in  this  way  what  hori- 
zontal force  is  employed  in  moving  the  sled  along  the  ground  and  also 
what  force  tends  to  lift  the  sled  vertically. 

3.  In  moving  a  weight  on  a  sled  as  in  the  previous  exercise,  it  was 
found  that  the  vertical  pull  was  40  pounds.  Determine  the  pull  on  the 
traces  by  drawing  to  scale.  What  is  the  horizontal  component  which 
moves  the  sled? 

What  is  the  effect  of  a  "long  hitch"  in  a  case  of  this  kind? 

4.  A  weight  of  40  lb.  slides  without  friction  down  an  incline  which 
makes  an  angle  of  30°  with  the  horizontal.  What  force  acting  directly 
up  the  plane  will  keep  the  weight  at  rest  ? 

[Hint.     Draw  as  in  the  accompanying  figure  the  segment  AB  40 
units  in  length  and  directly  downward  to  represent  the  force  due  to 
the  weight.     Project  this  segment  on  the  in- 
cline and  call  this  projection  AC.     This  is  the 
component  of  the  force  down  the  plane.     Now 
ZP=ZB  =  SO°  since  S^ABC,  PQR  are  similar. 
But  ABC  is  half  of  an  equilateral  triangle  and 
the  side  AC  is  one  half  oi  AB.     Thus  AC  rep- 
resents a  force  of  20  lb.     Therefore  a  force  of  Pj^    ^^ 
20  lb.  acting  directly  up  the  plane  will  keep 

the  weight  at  rest.  The  pressure  on  the  incline  is  represented  by  CB 
perpendicular  to  it.     Find  it.] 

5.  What  force  is  necessary  to  roll  a  barrel  of  cider  weighing  200  lb. 
up  an  incline  12  ft.  long  into  a  wagon  bed  3  ft.  above  the  ground? 


CHAPTER  II 
COMPUTATION  —  MEASUREMENT 

6.  Measurements  Are  Approximations.  Numbers  met 
with  in  actual  life  are  in  general  approximations.  For 
example,  if  we  should  measure  the  width  of  a  room  and  find 
the  measurement  to  be  13'  6f ",  obviously  we  could  not  be 
certain  that  this  is  the  exact  width.  A  more  exact  measure- 
ment could  be  obtained  by  using  a  ruler  graduated  to  smaller 
divisions,  for  example  to  sixty-fourths  of  an  inch.  The 
width  of  the  room  is  thus  not  measurable  to  an  exact  fraction 
of  an  inch.  How  nearly  the  measured  width  approximates 
the  true  width  depends  upon  the  wishes  of  the  operator,  upon 
his  skill,  upon  the  kind  of  instrument  used,  etc.  Even  the 
price  of  butter  at  67^  per  pound,  while  seemingly  exact,  is 
only  approximate,  if  the  weighing  process  be  considered. 

Again,  common  fractions  whose  denominators  contain 
factors  other  than  2  or  5  cannot  be  expressed  exactly  as 
terminating  decimals.  For  example,  one  sixth  of  a  dollar 
is  16.66  •  *  •  ff  =  17f^.  The  value  may  be  expressed  decimally 
to  any  degree  of  accuracy  desired.  Roots  of  numbers,  while 
exact  in  themselves,  are  in  general  only  approximations  when 
expressed  in  decimal  form. 

Thus,  1/6  =  0.17,  or  =  0.167,  or  0.1667,  etc.,  and  V2  =  1.4, 
or  =1.414,  or  =1.4142,  depending  upon  the  degree  of  accu- 
racy required.  In  all  cases  the  number  should  express  only 
the  degree  of  accuracy  demanded  by  the  nature  of  the  prob- 
lem. The  cost  of  food  to  $0,001,  or  the  volume  of  a  barn  to 
1  cu.  ft.,  are  meaningless. 

10 


II,  §  7]         COMPUTATION  —  MEASUREMENT  11 

7.   Accuracy  of  Processes  Involving  Approximations.    The 

accuracy  or  precision  of  a  measurement  is  indicated  by  the 
number  of  significant  *  figures  used  to  express  the  result  and 
not  by  the  number  of  decimal  places.  For  example,  52.3 
inches  =  4.83  feet  =1.35  meters  =  .293  rod  according  to  the 
unit  used.  Notice  that  these  four  members  represent  the 
same  length  and  are  about  equally  accurate.  They  each 
contain  three  significant  figures  but  in  one  case  one  decimal 
place  is  used,  in  another  three  decimal  places.  These 
numbers,  and  others  like  them,  are  said  to  be  of  three-figure 
accuracy.  Such  numbers  may  be  obtained  in  measurements 
made  with  ordinary  instruments  and  skill. 

Four-figure  accuracy  is  sufficient  for  a  great  many  engineer- 
ing operations.  Five-figure  accuracy  results  only  from  good 
instruments  and  great  care.  Thus  a  lot  on  Broadway,  New 
York  City,  is  measured  to  thousandths  of  a  foot.  Only 
experts  can  obtain  six-figure  accuracy,  while  eight-figure 
accuracy  is  as  yet  obtainable  only  in  rare  instances. 

In  working  with  approximate  numbers,  the  accuracy  of 
calculated  results  can  seldom  equal  that  of  the  original 
numbers  used.  If  a  measured  quantity  is  expressed  in  three 
figures,  the  result  of  operating  with  it  should  be  expressed 
in  general  in  not  more  than  three  figures  by  merely  dropping 
unnecessary  decimals.  For  example,  2.36  feet  =  2.36X12 
inches  =  28.3  inches,  not  28.32  inches.  Here  2.36  feet  as  a 
measurement  means  that  the  true  result  is  somewhere 
between  2.355  and  2.365.     This  is  written  (2.36 ±.005)  feet. 


*  All  figures,  other  than  zero,  are  significant  in  a  number.  A  zero  is 
significant  unless  used  merely  to  locate  the  decimal  point.  In  206,  2.06, 
0.206  the  central  zero  is  significant  and  these  are  called  three-figured  num- 
bers. In  0.0003  the  zeros  are  not  significant  and  this  is  called  a  one-figured 
number.  A  final  zero  may  be  significant ;  thus  2.60  ft.  means  that  the 
length  measured  is  nearer  2.00  ft.  than  to  2.59  ft.  or  to  2.61  ft.  If  the 
result  were  written  2.6  ft.,  it  would  mean  that  the  measurement  had  been 
made  only  to  the  nearest  tenth  of  a  foot. 


12  MATHEMATICS  [II,  §  7 

Now  (2.36  ±0.005)  12  inches  =  (28.32  ±0.06)  inches,  which 
shows  that  the  final  figure  is  in  doubt.  It  should  therefore 
be  written  28.3  inches  if  no  accompanying  statement  is 
made  as  to  the  error  involved. 

However,  it  is  usual  to  determine  the  doubtful  place  in  a 
result  from  accompanying  statements  of  probable  error  in 
the  numbers  used,  and  then  drop  unnecessary  figures  beyond 
the  first  doubtful  one. 

8.  Percentage.  The  word  per  cent  is  a  contraction  of  the 
Latin  word  per  centum  and  means  by  the  hundred,  i.e. 
hundredths.  It  is  written  %.  It  furnishes  a  convenient 
way  of  making  comparisons  between  quantities,  rates,  etc., 
by  using  a  hundred  as  the  basis.  Thus, 
6%  means  .06  or  ^^ 
135%  means  1.35  or  f^ 

Example.     An  analysis  of  a  pound  of  alfalfa  seed  showed 
weed  seed,  2^^  oz. ;   dirt,  Jf  oz. ;   good  seed,  12^  oz. 
The  relative  amounts  of  weed  seed,  dirt,  and  good  seed  may  be  more 
easily  recognized  by  stating  them  in  hundredths  of  a  pound,  i.e.  in 
per  cents.     The  analysis  thus  shows 

weed  seed  2^^  oz.  =  .16  lb.  =  16% 

dirtif  oz.  =  .06  1b.=6% 
good  seed  12|  oz.  =  .78  lb.  =78% 

Computations  involving  per  cent  contain  the  following 
three  elements. 

The  Rate  per  cent,  which  is  the  number  of  hundredths  taken. 

The  Base,  which  is  the  number  on  which  the  hundredths 
are  computed. 

The  Percentage,  which  is  the  result  of  taking  a  certain 
per  cent  of  a  number. 

Rules.     Base  times  Rate  =  Percentage. 

Percentage  divided  hy  Base  =  Rate. 
Percentage  divided  hy  Rate  =  Base. 


II,  §  8]  COMPUTATION  —  MEASUREMENT  13 

Example  1.     Find  8%  of  $150. 

8%  of  $150  =  .08  times  $150  =  $12.00 
Example  2.     What  per  cent  of  750  is  25? 

25  is  ^%%  of  750,  but  y^^j,  =^^0  =  -031  =3^%. 
Example  3.     $1.05  is  15%  of  what  amount? 

First  Solution  :  By  the  rule,  the  amount  required  or  the  base 
=  $1.05-^.15  =  $7.00 

Second  Solution:   If  15%  of  some  number  =  $1.05,  then  1%  of 
that  number  =  $.07,  and  100%,  or  the  number,  =$7.00 
Third  Solution  :   Let  x  be  the  amount  required.    Then 
.15x  =  $1.05, 

whence  a;  =  ^^^^  =  $7.00 

.15 

The  following  equivalents  may  be  found  useful  in  esti- 
mating and  in  computing : 

6%   =.06  =  ^.  m%  =  i.  62i%  =  f. 

6i%  =  .06i  =  A.  16|%  =  i.  66f  %  =  f . 

6f%  =  iV.  25%   =i.  83i%  =  f 

7|%  =  A.  33i%  =  f  87i%  =  |. 

8i%  =  iV.  37i%  =  |. 

EXERCISES 

1.  Find  6i%  of  300.  Of  75.  Of  18.  Find  23%  of  1684.  |%  of 
$24.     31%  of  180  lb. 

2.  What  per  cent  of  15  is  8?  What  per  cent  of  20  is  9?  What  per 
cent  of  12|  is  3  ?     What  per  cent  of  3|  is  2|  ?     What  per  cent  of  8  is  10 ? 

3.  What  per  cent  of  371  feet  is  5  feet?  What  per  cent  of  7  yards 
is  7  feet?  What  per  cent  of  2  miles  is  528  feet?  What  per  cent  pure 
is  an  18-karat  gold  ring? 

4.  6  is  8%  of  what  number?  4  is  10%  of  what?  18  is  2^%  of 
what?  24  is  12%  of  what?  216  is  13%  of  what?  6  feet  2.8  inches  is 
22%  of  what  length?     9.9  inches  is  12^%  of  how  many  yards? 

5.  In  a  certain  fertilizer  for  corn  land,  47^%  is  cotton-seed  meal, 
43|%  is  phosphoric  acid  (phosphorus),  and  the  rest  kainit.  How  many 
pounds  of  each  are  there  in  a  ton  of  the  fertilizer? 

How  much  acid  phosphate  containing  16%  phosphoric  acid  is  re- 
quired to  supply  the  phosphoric  acid? 


14  MATHEMATICS  [II,  §  8 

How  much  potash  is  there  in  the  kainit  required  if  kainit  contains 
13.54%  potash? 

6.  Cooked  eggs  contain  12%  protein  and  cooked  roast  beef  contains 
22.4%  protein.  If  one  egg  weighs  2  oz.,  how  many  eggs  will  furnish 
as  much  protein  as  1  pound  of  roast  beef  ?  Which  is  the  more  economi- 
cal food  with  eggs  at  Q5^  per  dozen  and  beef  at  25^  per  pound? 

7.  Ear  corn  shrinks  in  weight  about  3%  per  month  in  drying.  On 
Oct.  1  a  man  has  20  tons  of  corn.  What  should  it  weigh  the  following 
March  15?  Which  is  the  better  proposition,  to  sell  it  "green"  at 
$30  per  ton  or  hold  it  till  March  15  and  sell  it  for  $35  per  ton,  allowing 
6%  interest  on  the  value  of  the  corn?  What  "present"  selling  price 
(October  1)  at  6%  interest  would  be  the  equivalent  of  $35  per  ton 
after  drying  (March  15)  ? 

8.  Workmen  strike  for  a  10%  raise  on  their  present  scale  of  $2.25 
per  day.  If  they  "stay  out"  50  days,  how  long  will  it  take  them  under 
the  new  scale  to  regain  their  loss,  if  they  work  285  days  to  the  year, 
not  considering  interest  on  the  money  involved  ? 

9.  The  champion  Jersey  cow  (1914)  (Sophie  19th  of  Hood  Farm) 
produced  999.14  pounds  of  butter  fat  in  365  days.  Her  milk  tested 
5.69%  butter  fat. 

The  champion  Guernsey  cow  of  the  world,  Murne  Cowan  (for  a 
time  champion  of  all  breeds,  1915),  produced  24,008  pounds  of  milk  in 
365  days,  testing  4.57%  butter  fat. 

When  Finderne  Pride  Johanna  Rue  (Holstein)  completed  her  record 
in  June,  1915,  she  became  the  world's  champion  dairy  cow.  She  pro- 
duced, in  365  days,  28,403.7  pounds  of  milk  containing  1176.47  pounds 
of  butter  fat. 

Compare  these  three  breeds  of  dairy  cows  on  the  basis  of  the  amount 
of  milk  produced ;  on  the  basis  of  the  amount  of  butter  fat  produced ; 
on  the  basis  of  the  percentage  of  butter  fat  in  the  milk. 

9.  Trade  Discounts.  Profit  and  Loss.  Merchants, 
jobbers,  manufacturers,  publishers,  etc.  usually  have  a  fixed 
price,  called  the  list  price,  from  which  they  allow  discounts 
to  the  trade,  according  to  the  credit  of  the  customer,  the 
amount  purchased,  the  time  of  payment,  etc.  If  there  is 
more  than  one  discount,  it  is  understood  that  the  first  is  the 


¥ 


II,  §  9]  COMPUTATION  —  MEASUREMENT  15 

discount  from  the  list  price,  the  second  is  the  discount  from 
the  remainder,  etc. 

The  per  cent  of  profit  or  loss  in  a  transaction  is  usually 
reckoned  on  the  cost.  For  example,  if  an  article  costing  $6  is 
sold  for  $7.20,  the  gain  is  $1.20,  i.e.  it  is  ^  or  20%  of  the  cost. 

EXERCISES 

1.  What  is  the  cash  cost  of  12  bales  of  barb  wire  at  $3.75,  subject 
to  a  10%  discount,  and  a  further  discount  of  2%  if  paid  in  10  days? 

2.  Supphes  for  fences  amounted  to  $58.75.  If  a  discount  of  12|% 
is  allowed  with  an  additional  2%  off  for  cash,  what  is  the  cash  cost  of 
the  supplies? 

3.  A  dealer  is  offered  40%  and  15%  off  on  a  bill  of  $126.83.  What 
is  the  net  price  ? 

4.  Which  is  the  better  proposition,  to  pay  a  bill  according  to  a 
straight  discount  of  25%,  or  according  to  successive  discounts  of  15%, 
5%,  5%?     How  much  better  is  it? 

5.  A  dealer  buys  apples  at  $6.75  per  barrel  (3  bu.)  and  retails  them 
at  75^  per  peck.  What  is  his  gain  or  loss  on  30  barrels,  allowing  10% 
loss  by  decay  ? 

6.  A  renter  raises  60  bu.  per  acre  on  40  acres  of  corn  where  land 
is  valued  at  $180  per  acre.  He  sells  the  corn  at  $1.25  per  bushel,  pays 
a  rental  of  8%  of  the  farm  value,  and  estimates  the  cost  of  crop  produc- 
tion at  $45.75  per  acre.  How  much  and  what  per  cent  profit  does  he 
make  on  the  cost  of  production? 

7.  A  quantity  of  wheat  was  sold  in  succession  by  three  dealers, 
each  of  whom  made  a  profit  of  4%.  The  last  dealer  received  $1384, 
What  did  it  cost  the  first  dealer? 

8.  What  per  cent  profit  does  a  grocer  make  on  sugar  bought  at 
14^^  per  pound,  depreciated  10%  from  drying  and  downweight,  and 
sold  at  16)!f  per  pound? 

9.  A  man  buys  2000  bu.  of  potatoes  at  $1.45,  sells  f  of  them  at  20% 
gain  and  the  remainder  at  a  loss  of  1H%.      Find  his  net  rate  of  gain. 

10.  Goods  are  bought  at  10%  below  list  price  and  sold  at  20%  above 
list  price.     What  is  the  per  cent  profit? 

11.  If  a  dealer  gets  discounts  of  40%  and  15%,  what  discount  can 
he  give  on  the  list  price  to  make  30%  on  his  investment? 


16  MATHEMATICS  [11,  §  10 

10.  Simple  Interest.  Interest  is  money  paid  for  the  use 
of  mone^^  The  principal  is  the  amount  borrowed.  The 
rate  is  the  part  of  the  principal  to  be  paid  for  its  use,  usually 
for  one  year,  even  though  it  may  be  collected  semiannually 
or  quarterly  if  so  specified.  Thus  the  statement  that  a  note 
or  a  bond  is  issued  at  6%  interest  payable  semiannually 
means  that  interest  is  payable  semiannually  at  the  rate  of 
6%  per  year.  In  computing  simple  interest,  thirty  days  are 
considered  an  interest  month  and  360  days  an  interest  year. 
The  United  States  Government  and  some  banks  and  business 
houses  use  the  exact  interest  method.  This  method  takes 
account  of  the  exact  number  of  days  in  the  time  interval,  and 
proceeds  upon  the  basis  of  365  days  to  the  year.  Banks 
usually  compute  interest  for  the  exact  number  of  days. 

11.  Partial  Pa3mients.  Sometimes  it  is  convenient  to  pay 
part  of  a  note  before  it  matures.  Such  payments  should  be 
indorsed  on  the  back  of  the  note,  with  the  date  and  the 
amount  of  the  payment.  They  are  known  as  partial  pay- 
ments. 

The  Supreme  Court  of  the  United  States,  and  nearly  all 
of  the  states,  use  the  following  rule  for  partial  payments, 
called  the  United  States  Rule, 

1.  Find  the  amount  of  the  principal  to  the  time  of  the  first 
payment;  if  the  payment  equals  or  exceeds  the  interest,  subtract 
the  payment  from  the  amount  and  treat  the  remainder  as  a  new 
principal. 

2.  If  the  payment  is  less  than  the  interest,  find  the  amount 
of  the  same  principal  to  the  time  when  the  sum  of  the  payments 
shall  equal  or  exceed  the  interest  due,  and  subtract  the  sum  of 
the  payments  from  the  amount. 

3.  Proceed  in  the  same  manner  with  the  remaining  payments 
until  the  time  of  settlement. 


II,  §  12]        COMPUTATION  —  MEASUREMENT  17 

EXERCISES 

1.  $304.75  Chicago,  111.,  March  10,  1912. 
One  year  after  date,  for  value  received,  I  promise  to  pay  to  G.  Botha, 

or  order,  Three  Hundred  Four  and  ^^q  Dollars,  with  interest  at  6%. 

B.  Thoba. 
Indorsements :   July  12,  1912,  $100 ;   Nov.  30,  1912,  $100. 
What  was  due  June  25,  1914? 

2.  $429.30  Columbus,  O.,  April  13,  1913. 
On  demand,  I  promise  to  pay  Spencer  and  Arps,  or  order,  Four  Hun- 
dred Twenty-nine  and  y^/j  Dollars,  value  received,  with  interest  at  6%. 

R.  Macklin. 
Indorsements:  Oct.  2,  1913,  $10;    Dec.  14,  1914,  $120;    July  1, 
1915,  $25. 

What  was  due  Jan.  1,  1916? 

3.  A  note  for  $650,  at  5%,  is  dated  June  1,  1907,  and  has  the  follow- 
ing partial  payments  indorsed  on  it :  Aug.  2,  1907,  by  labor,  $5.00 ; 
Aug.  8,  1907,  by  labor,  $2.00;  Oct.  1,  1907,  $95;  June  1,  1908,  $275; 
Dec.  1,  1908,  $140.     How  much  is  due  Jan.  1,  1909? 

12.  Present  Worth.  Bank  Discoimt.  The  present  worth 
of  k  dollars,  payable  at  a  given  future  time,  is  the  sum  that 
will  amount  to  k  dollars  at  the  end  of  the  time,  at  a  specified 
rate  of  interest. 

The  interest  on  the  present  worth  of  a  debt  due  at  a  future 
time  is  called  true  discount.  It  is  the  difference  between 
the  debt  and  its  present  worth. 

Example.  What  is  the  present  worth  of  $318  due  in  1  year  if  money 
is  worth  6%  ? 

Solution  :  If  x  is  the  present  worth,  then  the  amount  of  x  for  1  30*. 
at  6%  is  equal  to  $318,  i.e. 

x(1.06)=$318, 

In  any  case,  we  have  the  following  rule : 

Rule  for  Present  Worth.  To  find  the  present  worthy 
divide  the  given  sum  by  the  amount  0/  $1.00  for  the  given  time 
at  the  given  rate. 


18  MATHEMATICS  [II,  §  13 

13.  Bank  Discount.  When  a  loan  is  made  at  a  bank, 
the  interest  is  paid  in  advance,  and  the  interest  thus  paid  is 
called  bank  discount. 

The  holder  of  a  note  which  is  due  at  a  certain  time  may- 
want  the  value  of  the  note  before  it  is  due.  In  this  case  he 
usually  takes  it  to  a  bank  where  the  note  is  discounted,  i.e. 
the  interest  on  the  full  value  of  the  note  at  maturity,  from  the 
time  of  discount  to  the  time  of  maturity,  is  subtracted  from  the 
amount  of  the  note,  and  the  remainder,  called  the  proceeds,  is 
paid  to  the  holder  of  the  note.  The  note  then  becomes  the 
property  of  the  bank. 

Example.  A  note  of  $600,  dated  March  1,  1908,  interest  6%,  to 
run  4  months,  was  discounted  at  6%,  March  20,  1908.  Find  the  pro- 
ceeds. 

Amount  of  $600  for  4  mo.  @  6%  $612.00 

Interest  on  $612  @  6%  for  105  days,  March  20  to  July  3, 

(called  the  bank  discount)  10.71 

Proceeds,  after  deducting  the  bank  discount,  $601.29 

EXERCISES 

1.  What  is  the  present  worth  of  $460.50  due  in  3  yr.  9  mo.  18  days, 
if  money  is  worth  6% ?  Ans.    $375. 

2.  Find  the  present  worth  of  $375  due  in  3  yr.  3  mo.  at  6%. 

Ans.     $313.81 

3.  Which  is  worth  most,  $320  in  12  mo.,  $310  in  6  mo.,  or  $300  cash, 
if  money  is  worth  6% ?  Ans.     The  first. 

4.  A  note  of  $700  dated  Aug.  9,  1915,  is  due  Feb.  12,  1916,  and 
bears  6%  interest.  What  is  the  present  worth  of  this  note  on  Oct.  8, 
1915?  What  are  the  proceeds  on  that  date  if  discounted  in  a  bank 
at  7%? 

5.  A  man  buys  a  farm  on  March  25,  1911,  agreeing  to  pay  $2700, 
Oct.  1,  1911,  and  $2500,  Jan.  1,  1912.  If  a  discount  of  7%  is  allowed 
for  cash,  how  much  will  he  gain  by  borrowing  the  money  on  his  own 
note  discounted  at  6%?     How  much  must  he  borrow? 


II,  §  14]        COMPUTATION  —  MEASUREMENT  19 

6.  A  note  of  $560  to  run  3  months,  dated  Aug.  4,  1914,  bearing  5% 
interest,  was  discounted  at  6%  on  Sept.  19.  Find  the  proceeds  of  this 
note. 

7.  A  man  sold  his  farm  for  $6350,  taking  in  payment  a  note  at 
5%  interest  due  in  6  mo.  He  at  once  sold  the  note  to  a  bank  which 
discounted  it  at  6%.  What  were  the  proceeds,  i.e.  what  did  he  get 
in  cash  for  his  farm  ? 

8.  A  note  of  $350  due  in  one  year  at  6%  was  discounted  54  days 
before  it  was  due  at  6%.     What  were  the  proceeds  of  the  note? 

14.  Elementary  Mensuration.  Angles.  In  the  following 
pages  we  shall  treat  angles,  the  lengths  of  Hnes  and  their 
ratios,  the  areas  of  surfaces,  and  the  volumes  of  solids. 

If  two  straight  lines  proceed  from  the  same  point  they  form 
a  plane  angle.  This  point  is  the  vertex  of  the  angle.  If 
there  is  only  one  angle  at  a  vertex,  the  angle 
may  be  denoted  by  a  capital  letter  at  the  ver- 
tex, as  ZO,  Fig.  15.  Any  angle  may  be  de- 
noted by  three  letters,  as  ZAOPj  Fig.  15, 
the  middle  letter  being  the  vertex. 

An  angle  may  be  thought  of  as  generated  by  a  line  turning 
about  its  end-point  as  a  pivot.  Thus,  if  the  ray  OP,  Fig.  16, 
revolves  about  0,  the  amount  of  the 
turn  from  OX  to  OP  measures  the  angle 
XOP  indicated  by  the  arrow.  , 

If  the  ray  continues  to  rotate  counter- 
clockwise  until  it  is  exactly  opposite  its 
original  position  it  generates  a  straight  ' 

angle,  as  XOX' ,  Fig.  16.  If  the  ray  continues  till  it  reaches 
its  original  position,  it  generates  a  complete  rotation,  or  one 
revolution. 

A  unit  of  measure  for  angles  is  the  degree.  A  complete 
rotation  is  360  degrees,  written  360°.  Let  the  student  state 
the  definitions  of  right  angle,  acute  angle,  obtuse  angle. 


20 


MATHEMATICS 


[II,  §  15 


15.   Triangles.     Let  the  student  give  definitions  of  triangle, 
equilateral  triangle,  isosceles  triangle,  right  triangle  (Fig.  17). 

The  sum  of  the  three  angles  of  any  triangle  is  two  right  angles, 
or  180°. 


Oblique 

Triangle 


Right 
Triangle 


Fig.  17 


It  is  customary,  as  shown  in  Fig.  17,  to  designate  the  sides 
of  a  triangle  by  the  three  small  letters  a,  6,  c  and  the 
angles  opposite  these  sides  by  the 
corresponding  capital  letters.  A,  B, 
C,  respectively. 

In  a  right  triangle  the  square  of 
the  hypotenuse  equals  the  sum  of  the 
squares  of  the  other  two  sides,  i.e. 
from  Fig.  18, 


a'-\-b\ 


'c      a  _ 

b 

1      I       C 


di  =  a^  +  l^ 
Fig.  18 


This  fact  is  called  the  Pythagoras 

theorem.* 

When  the  base  and  the  altitude  of  a  triangle  are  given 

the  area  is  j 

A  =  ^base  X  altitude. 

When  the  three  sides  are  given  the  area  is 
A  =  \/s  {s  —  a){s  —  b)  {s  —  c). 


*  Tables  of  squares  and  square  roots  are  found  in  the  Appendix.     They 
may  prove  of  assistance  in  the  computation. 


II,  §  15]        COMPUTATION  —  MEASUREMENT  21 

where  a,  b,  and  c  denote  the  three  sides,  respectively,  and 

s  =  i  (a+b+c). 

State  in  words  the  formula  for  the  area  of  a  triangle  when 
the  three  sides  are  given. 

Example.     What  is  the  area  of  a  triangle  whose  sides  are  15.3,  16.4, 
and  17.7  rods. 

Solution  :   Let  a  =  15.3,  h  =  16.4,  c  =  17.7 ;   then 

s±=i(a+64-c)=24.7,  s-a  =  9.4,  s-6  =  8.3,  s-c  =  7.0 
Therefore  A  =  V(24.7)(9.4)(8.3)(7.0)  =  116.2  square  rods. 


EXERCISES 

1.  A  ladder  16  ft.  long  and  standing  on  level  ground  just  reaches 
a  window  12  ft.  from  the  ground.  Assuming  the  wall  to  be  perpen- 
dicular, how  far  from  the  base  of  the  wall  is  the  foot  of  the  ladder  ? 

2.  Find  the  length  of  the  hypotenuse  of  a  right  triangle  whose  other 
two  sides  are  equal  and  whose  area  is  two  acres. 

How  many  degrees  are  there  in  each  of  the  acute  angles  of  this 
triangle  ? 

3.  Find  the  area  of  a  triangle  whose  sides  are  13,  18,  and  21  in. 

4.  Find  the  altitude  and  the  area  of  an  isosceles  triangle  whose 
equal  sides  are  8  ft.  long,  and  whose  base  is  6  ft.  long. 

5.  Find  the  length  of  the  longest  umbrella  that  can  be  put  di- 
agonally in  a  trunk  which  measures  20  X  20  X  34  inches. 

6.  The  dimensions  of  a  barn  are  20'  X  44'  X  48'.  What  is  the 
length  of  a  wire  which  connects  the  upper  corner  at  the  eaves  to  the 
opposite  lower  corner? 

7.  A  house  is  located  10  rods  from  a  certain  river,  and  a  barn  is 
15  rods  from  this  house  and  19  rods  from  the  same  river.  Find  the 
point  on  the  river  equidistant  from  the  house  and  the  barn  and  find 
this  distance. 

8.  A  farmer  wishes  to  run  a  pipe  from  some  point  on  the  river  of 
Ex.  7  to  the  house  and  another  pipe  from  the  same  point  on  the  river 
to  the  barn.  Where  should  it  be  laid  to  require  the  least  length  of 
pipe?    Find  X\m  length. 


22 


MATHEMATICS 


[II,  §  16 


16.    Quadrilaterals.     Let  the  student  define  quadrilateral, 
square,  rectangle,  parallelogram,  trapezoid.     (See  Fig.  19.) 


Square 


Rectangle 


Trapezoid 


Parallelogram 


Fig.  19 

The  areas  of  such  quadrilaterals  may  be  found  by  the 
following  rules. 

1 .  Square : 
Area  =  s^, 

where  s  =  the  length  of  one  side. 

2.  Rectangle  or  Parallelogram: 
Area  =  base  X  altitude. 

3.  Trapezoid  : 

Area  =  altitudeXhalf  the  sum  of  the  parallel  sides,  i.e. 
_   (6+c) 
-"    2     ' 
where  a  is  the  altitude,  and  h  and  c  are  the  parallel  sides. 

EXERCISES 

1.  State  in  words  the  rule  for  the  area  of  a  trapezoid. 

2.  Find  the  area  of  a  trapezoidal  field  whose  parallel  sides  are 
10  rods  and  16  rods  long  and  the  distance  between  them  is  8  rods. 

3.  Figure  20  as  dimensioned  represents  the 
gable  end  of  a  building.  Compute  the  length 
of  the  rafter  OR  and  the  area  of  ABDOC. 
What  is  the  rise  of  the  rafter  per  foot  of  its 
run?    What  is  the  pitch  of  the  roof?    (See  §  4.) 

4.  Show  how   a    carpenter    ''lays    off"   a  t?t      90 
rafter  by  use  of  a  steel  square,  i.e.  how  he  cuts 

the  desired  bevel  on  the  rafter,  making  use  of  the  ratio  of  rise  to  run. 


II,  §  17]        COMPUTATION  —  MEASUREMENT 


23 


5.  Compute  the  pitch  and  the  length  of  each 
rafter  and  the  area  of  the  entire  gable  end  of  the 
building  as  shown  in  Fig.  21. 

6.  ABC  is  a  triangle,  two  of  whose  sides  are 
AC  =  17'  and  BC  =  10'.  The  projection  of  BC  on 
the  base  AB  i^Q'.  Draw  the  several  cases  possi- 
ble here  and  find  the  area  of  ABC  for  each  case. 


Fig.  21 


17.  Polygons.  Any  plane  figure  bounded  by  straight 
lines  is  called  a  polygon.  The  sum  of  its  sides  is  called  the 
perimeter  of  the  polygon. 

A  polygon  of  five  sides  is  called  a  pentagon]  one  of  six 
sides,  a  hexagon ;  one  of  eight  sides,  an  octagon ;  one  of  ten 
sides,  a  decagon,  etc. 

A  regular  polygon  is  one  whose  sides  are  all  equal,  and 
whose  angles,  between  adjacent  sides,  are  all  equal. 

The  center  of  a  regular  polygon  is 
the  common  center  of  its  inscribed 
and  circumscribed  circles  (§  19). 

A  regular  polygon  can  be  divided 
into  as  many  equal  triangles  as  the 
polygon  has  sides.  Each  such  tri- 
angle has  one  vertex  at  the  center 
of  the  polygon,  and  its  altitude, 
called  the  apothem  of  the  polygon, 
is  the  distance  from  the  center  of 
the  polygon  to  the  center  of  a  side,  as  OP,  Fig.  22. 

An  interior  angle  of  a  regular  polygon,  as  B,  Fig.  22,  is 
equal  to  (n  — 2)180° -j-n  where  n  is  the  number  of  sides  of  the 
polygon.  Thus,  for  the  regular  pentagon,  n  =  5,  and  hence 
B=(5-2)180°-^5  =  108°. 

The  area  of  a  regular  polygon  is 


EPA 
Regulab  Poltoons 

Fig.  22 


apothemXthe  perimeter. 


24  MATHEMATICS  [II,  §  17 

EXERCISES 

1.  Find  the  area  of  a  regular  hexagon  whose  side  is  1". 

2.  The  side  of  a  regular  decagon  is  1"  and  its  apothem  is  1.54" 
Find  its  area. 

3.  Prove  the  statement  in  the  text  that  the  interior  angle  of  a 
regular  polygon  is  equal  to  (n— 2)180° -^n,  by  drawing  the  equal  tri- 
angles AOE,  EOD,  etc.  as  in  Fig.  22,  and  taking  the  sum  of  their  in- 
terior angles  less  the  angles  at  the  center. 

4.  Find  the  number  of  degrees  in  an  interior  angle  of  a  regular 
hexagon.  A  regular  octagon.  A  regular  decagon.  A  regular  48-gon. 
A  regular  500-gon. 

18.  Irregular  Areas.  I.  Straight  Boundaries.  If  a  tract 
is  bounded  by  straight  lines,  i.e.  if  its  perimeter  is  a  polygon 
(Figs.  24-26),  the  area  may  be  computed  by  projecting  the 
sides  of  the  polygon  upon  some  convenient  base  line.  The 
resulting  trapezoids  may  be  combined  to  obtain  the  required 
area. 

The  area  can  also  be  computed  by  dividing  the  tract  into 
triangles  and  measuring  a  sufficient  number  of  dimensions  of 
each  triangle. 

II.  Curved  Boundaries.  A  curved  boundary  occurs 
when  a  tract  touches  a  stream  or  a  curved  road.  Stakes  are 
set  at  points  along  the  curved  part,  so  that  it  will  approximate 
closely  to  straight  lines  between  the  stakes.  The  perimeter 
then  takes  the  form  of  an  ^ 

irregular  polygon  and  may  f\    ^^^>v  ^       „ 

be  treated  accordingly.  !     j  |   ^v^d^/  |        \ 

For  example,  in  Fig.  23,  |     j  I  '  I        ! 

stakes  are  set  Sit  A,  B,  C,        p    Q        R        s         t       u 
etc.,  along  the  boundary.  ^^^'  ^^ 

The  sides  AB,  BC,  etc.  are  now  projected  on  a  convenient  base 
line,  and  the  offsets  from  the  stakes  to  the  base  line  are  meas- 
ured. The  areas  of  the  resulting  trapezoids  are  then  computed. 


II,  §  18]         COMPUTATION  —  MEASUREMENT 


25 


When  the  curved  boundary  is  not  too  abrupt  in  any  place, 
stakes  may  be  set  at  regular  intervals  on  the  base  Hne,  and 
offsets  to  the  curved  boundary  measured.  The  computation 
leads  to  the  following  general  rule  for  offsets  at  regular 
intervals  on  the  base  line. 

Rule.  Add  all  the  offsets  and  subtract  from  this  sum  half 
the  sum  of  the  extreme  offsets;  multiply  the  remainder  by  the 
common  distance  between  offsets. 

These  methods  form  the  basis  for  calculating  areas  in 
practical  surveying.  The  base  lines  in  surveying  are  usually 
the  magnetic  north-south  lines. 

EXERCISES 

1.  Find  the  area  of  a  triangular  field  ABC  in 
which  AE  =  5  rods;  EF  =  9  rods;  CE  =  10  rods; 
£F  =  7rods.     Fig.  24. 

[Hint.  ^^  is  the  base  line  and  AE  and  EF 
are  the  projections  of  AC  and  CB  respectively 
upon  it.  The  area  of  ABC  is  equal  to  the  right 
triangle  AEC  plus  the  trapezoid  CEFB  minus  the  right  triangle  AFB.] 

2.  Find  the  area  of  A  BCD  in  the  diagram,  cr 
taking 

PQ  =  20  rods       PD  =  40  rods       QR  =  10  rods 

QC  =  50  rods        RS  =  25  rods      RA^  15  rods 

^B  =  45  rods 

[Hint.  The  area  of  ABCD  =  PQCD+CQSB  - 
DPRA-ARSB,  Fig.  25. 

The  area  may  also  be  found  as  follows : 
parallel  to  PS  and  extend  PD  and  BS  to 
meet  it.  Take  the  area  of  the  outside  rec- 
tangle and  subtract  from  it  the  areas  of  the 
other  outside  figures.] 

3.  Find  the  area  of  the  five-sided  field 
ABCDE  shown  in  Fig.  26,  if 

xi/  =  5rods  Ax  =  15  rods 

Ey  =  18  rods  zu  =  15  rods 

uv  =  12  rods  Du  =  22  rods 


?-    p 

Q    R               S 

Fig.  25 

Draw 

a  line  through  C 

D 

5- 

^^\ 

[     j 

-^A^\ 

X   V 

Z              U             V 

Fig.  26 

yz  =  10  rods 

^2  =  4  rods 

Cy  =  llrods 

26 


MATHEMATICS 


[II,  §  18 


4.  In  surveying  a  field  between  the  road  AG  and  the  river  (Fig.  27), 
the  surveyor  measured  along  AG  and  drove  a  stake  every  100  feet, 
at  B,  C,  etc.  At  right  angles  to  the  road, 
the  distances  to  the  river  were  measured 
as  follows  :  From  B,  100  ft. ;  from  C,  220 
ft.;  from  D,  260  ft.;  from  E,  250  ft.; 
from  F,  230  ft.  Compute  the  approxi- 
mate area  of  the  field  and  its  value  at 
$190  per  acre. 

6.  Determine  the  area  between  a  base  line  and  an  irregular  boundary 
from  the  following  table  of  offsets  all  on  the  same  side  of  the  base  line. 
The  distances  are  measured  in  feet  between  successive  stakes  on  the 
base  line.     Give  the  result  in  square  feet,  and  in  acres. 


C    D    E    F    G 

Fig.  27 


Distances.     .     . 

0 

202 

156 

467 

159 

239 

Offsets.     .     .     . 

20.7 

31.5 

42.6 

53.2 

36.1 

40.7 

Ans.     49,698  sq.  ft. 
6.   The  following  offsets  were  taken  on  the  same  side  of  a  line  at 
distances  of  100  ft.  apart.     Find  the  areas  between  extreme  boundaries 
and  the  base  line. 


Distances     . 

0 

100 

200 

300 

400 

500 

600 

700 

-Offsets     .     . 

55.1 

76.1 

83.3 

79.9 

69.7 

59.5 

83.3 

70.5 

19.  Circles.  The  circle  occurs  very  frequently  in  applied 
mathematics,  perhaps  oftener  than  any  other  geometric 
form.  Let  the  student  state  the  definitions  of  circle,  center, 
radius,  diameter,  chord,  arc. 

The  area  bounded  by  a  chord  and  its  arc  is  called  a  seg- 
ment, as  AsB,  Fig.  28.  The  area  bounded  by  two  radii  and 
their  intercepted  arc  is  a  sector,  as  OAsBO,  Fig.  28. 

A  straight  line  which  touches  a  circle  at  only  one  point  is 
a  tangent,  as  TPS,  Fig.  28.  A  straight  line  which  intersects 
a  circle  in  two  points  is  a  secant,  as  CF,  Fig.  28. 


II,  §  20]         COMPUTATION  —  MEASUREMENT 


27 


A  radius  is  perpendicular  to  the  tangent  at  the  point  of  contact, 

A  polygon  is  inscribed  in  a  circle  when  its  sides  are  chords 

of  the  circle,  and  the  circle  is  then  circumscribed  about  the 


Fig.  28 

polygon.     A  circle  is  inscribed  in  a  polygon  when  each  of 
the  sides  of  the  polygon  is  tangent  to  the  circle. 

20.  Angles  of  a  Circle.  An  angle 
formed  by  two  radii  is  a  central 
angle,  as  AOB,  Fig.  28. 

An  angle  whose  vertex  lies  on  the 
circumference  of  a  circle  is  an  in- 
scribed angle,  as  QPR,  Fig.  28. 

I.  A  central  angle  is  said  to  be 
measured  by  the  arc  which  it  intercepts. 
This  means  that  a  given  central  angle  contains  a  number 
of  unit  angles  equal  to  the  number  of  unit  arcs  in  the  inter- 
cepted arc. 

II.  An  inscribed  angle  is  measured  by  one  half  of  its  inter- 
cepted arc.  Thus,  ZQPR  =  hixU  the  central  angle  whose 
arc  is  QR,  Fig.  28. 

III.  An  angle  between  a  tangent  and  a  chord  is  measured 
by  one  half  of  the  intercepted  arc.  Thus,  ZQPT  =  hsi\i  arc 
PQ,  Fig.  28. 


A         M         B 
FiQ.  29 


28  MATHEMATICS  [II,  §  21 

21.  Relations  between  Radius,  Arc,  Circumference  of  a 
Circle.  In  any  circle,  the  ratio  of  the  circumference  to  the 
diameter  is  3.1416,  approximately.  This  ratio  is  represented 
by  the  Greek  letter  tt  (pronounced  pi),  i.e. 

7r  =  3.1416  =  3i 
approximately.     Then 

Circumference  =  27rr, 
where  r  is  the  radius  of  the  circle. 

[Note.  The  exact  value  of  tt  cannot  be  computed.  Its  value  to 
the  first  ten  decimal  places  is 

TT  =  3.1415926535. 

Its  value  correct  to  over  700  decimal  places  has  been  computed.    As  a 
curiosity,  we  quote  the  value 

77  =  3.1415926535897932384626433832795028841971693993751.] 

The  length  of  any  arc  s  of  h  degrees  is  (/i/360)  •  27rr,  for  we 

have,  in  Fig.  28, 

Arc  g         ^  h° 
Circumference    360°  ^ 
hence 

<"'  ^  =  360  •  2'"'- 

Example  1.  Find  the  circumference  of  a  circle  whose  radius  is  5'. 
By  the  formula,  the  circumference  =  27r-  5'  =  31.4'. 

Example  2.  Find  the  length  of  the  arc  whose  central  angle  is 
30°  in  a  circle  whose  radius  is  5  ft. 

By  definition,  30°  of  arc  means  /^°o  of  the  whole  circumference. 
Then,  we  have 

arc  of  30°  =J^  X  27r  X  4  ft.  =5^  =2.62  ft. 
360  6 

22.  Area  of  a  Circle.  Area  of  a  Sector.  Area  of  a  Seg- 
ment. The  area  of  any  circle  is  equal  to  half  the  radius 
times  the  circumference,  i.e. 

area  of  circle  =  -  rX27rr  =  7rr^. 


II,  §  22]        COMPUTATION  —  MEASUREMENT  29 

The  area  of  a  sector  of  a  circle  whose  arc  is  h  degrees 
(Fig.  28)  is  /i/360  times  the  area  of  the  circle,  i.e. 

area  of  sector  =  ^^  irr'^. 

Example.     Find  the  area  of  a  sector  whose  arc  is  30°  in  a  circle 
whose  radius  is  5  ft. 

Area  of  sector  =^-T;7r  25=— j-^  =  6.55  sq.  ft. 

The  area  of  a  segment  of  a  circle,  as  ABsA,  Fig.  28,  is  the 
area  of  the  sector  OAsB  minus  the  area  of  the  triangle  OAB. 

EXERCISES 

1.  What  is  the  diameter  of  a  wagon  wheel  whose  circumference  is 
150  in.? 

2.  How  many  revolutions   per   second   is    a   30-inch   automobile 
wheel  making  when  the  speedometer  registers  25  mi.  per  hour? 

3.  What  is  the  area  of  the  cross  section  of  a  6-inch  stove  pipe?    A 
3-inch  one  ? 

4.  By  using  a  carpenter's  square  find  the  diameter  of  a  circular 
pipe  having  the  same  area  of  cross  section  as  the  sum  of  the  areas  of  a 
6-inch  and  an  8-inch  circular  pipe.  See  the  ac- 
companying figure.  Do  the  same  for  two  pipes 
whose  diameters  are  5|  in.  and  8y\  in.,  respectively. 

[Hint.     Use  the  theorem  that  similar  surfaces 

are  to  each  other  as  the  squares  of  their  like 

dimensions.] 

Fig.  30 

5.  Find  the  total  pressure  on  the  piston  of  a 

steam  engine  8  inches  in  diameter,  when  the  steam  gauge  registers  120 
pounds  per  square  inch. 

6.  A  circular  pond  has  a  diameter  of  75  yards.     What  is  the  cost 
at  $3  per  square  yard  of  constructing  a  walk  40"  wide  around  it  ? 

7.  WTiat  must  be  the  length  of  a  binding  rod  for  a  cylindrical 
silo  12  feet  in  diameter  allowing  1  ft.  for  overlapping? 

8.  Find  the  length  of  the  arc  and  the  area  of  the  corresponding 
sector  whose  central  angle  is  40°  in  a  circle  of  radius  10  inches. 


30 


MATHEMATICS 


[II,  §  22 


9.   Find  the  area  of  the  segment  of  a  circle  whose  arc  is  60°,  the 
radius  being  8  inches. 

[Hint.  The  side  opposite  the  30°  angle  in  a  right  triangle  whose 
acute  angles  are  30°,  60°  is  one  half  of  the  hypotenuse.] 

10.  The  central  angle  whose  arc  is  equal  to  the  radius  is  often  used  as 
the  unit  of  measure  of  angles.  It  is  called  a  radian.  Find  the  number 
of  degrees  in  a  radian. 

11.  Inscribe  an  equilateral  triangle  in  a  circle.  Show  the  relation 
between  the  angles  of  the  triangle  and  the  arcs  which  they  subtend. 

12.  Has  a  degree  of  arc  the  same  length  in  two  unequal  circles? 

13.  The  sizes  of  men's  hats  are  indicated  by  the  diameter  of  a  circle 
whose  perimeter  is  the  distance  around  the  head  where  the  hat  rests. 
What  size  of  hat  does  a  man  need  if  the  distance  around  his  head  meas- 
ures 22  inches  ? 

14.  If  you  wear  a  7\  iiat,  what  is  the  distance  around  your  head? 
Verify. 

23.  Polyhedrons.  A  polyhedron  is  a  solid  bounded  by 
planes.     A  polyhedron  of  four  faces  is  called  a  tetrahedron; 


ICOSAHEDRON        DODECAHEDRON       OcTAHEDRON 


CUBB 


Tetrahedron 


Fig-  31.  —  The  Five  Regular  Sc 


one  of  six  faces,  a  hexahedron;  one  of  eight  faces  an  octa- 
hedron; etc.  (Figs.  31,  32). 


II,  §  24]        COMPUTATION  —  MEASUREMENT 


31 


24.  Prisms.  A  prism  is  a  polyhedron,  two  of  whose  faces, 
called  its  bases,  are  equal  polygons  in  parallel  planes,  and 
whose  other  faces,  called  lateral  faces,  are  parallelograms 


Right  Prisms 


Oblique  Prisms 


Fig.  32 


whose   vertices   all   lie  in   the   bases.      The   altitude   of   a 
prism  is  the  perpendicular  distance  between  its  bases. 

A  prism  whose  bases  are  perpendic- 
ular to  its  lateral  edges  is  called  a  right 
prism  (Fig.  32).  Otherwise,  the  prism 
is  oblique. 

Prisms  are  called  triangular,  quad- 
rangular, etc.,  according  as  their  bases 
are  triangles,  quadrilaterals,  etc. 

A  right  section  of  a  prism  is  a  section 
perpendicular  to  the  lateral  edges  of  the 
prism,  as  ABODE,  Fig.  33. 

A  right  prism  is  regular  if  its  bases  are  regular  polygons. 


Right  Section 
Fig.  33 


32 


MATHEMATICS 


[II,  §  25 


25.   Parallelepipeds.     A  parallelepiped  is  a  prism  whose 
bases  are  parallelograms. 


Rectangular 
Parallelepiped 


Cube 


Oblique 
Parallelepiped 


Fig.  34 


A  rectangular  parallelepiped  is  one  whose  six  faces  are  all 
rectangles.  For  example,  most  boxes  are  rectangular  paral- 
lelepipeds. 

A  cube  is  a  parallelepiped  whose  six  faces  are  all  squares. 


26.   Unit  of  Volume.     The  unit  of  volume  is  a  cube  whose 
edges  are  equal  to  the  linear  unit  used  in  the  measurement. 

8 


2Z 


^ 


.7^0 


® 


Unit  of  Volume 
Fig.  35 


The  volume  of  any  solid  is  the  number  of  units  of  volume 
which  it  contains. 


II,  §  27]        COMPUTATION  —  MEASUREMENT 


33 


27.  P5rramids.  A  pyramid  is  a  polyhedron  of  which  one 
face,  called  the  base,  is  a  polygon  and  the  other  faces  are 
triangles  having  a  common  vertex,  called  the  vertex  of  the 
pyramid. 

The  altitude  of  a  pyramid  is  the  length  of  the  perpendicular 
let  fall  from  the  vertex  to  the  plane  of  the  base. 


B  C 

Regular  Pyramid 
Fig.  36 

A  pyramid  is  regular  if  its  base  is  a  regular  polygon  whose 
center  coincides  with  the  foot  of  the  perpendicular  let  fall 
from  the  vertex  to  the  base. 

The  slant  height  of  a  regular  pyramid  is  the  altitude  of  any 
one  of  the  lateral  faces,  as  VH,  Fig.  36. 


Frustum  of  a  Pyramid 
Fig.  37 


A  frustum  of  a  pyramid  is  the  portion  of  a  pyramid  in- 
cluded between  the  base  and  a  section  parallel  to  the  base. 

The  altitude  of  a  frustum  is  the  length  of  the  perpendic- 
ular between  the  planes  of  its  bases. 


34 


MATHEMATICS 


[II,  §  28 


28.  Cylinders.  A  cylindrical  surface  is  a  curved  surface 
generated  by  a  straight  line  which  moves  parallel  to  a  fixed 
straight  line  and  constantly  touches  a  fixed  curve  not  in  the 
plane  of  the  straight  line.     For  convenience  in  drawing,  as 


Cylindrical  Surface 


Right  Cylinder 
Fig.  38 


Oblique  Cylinder 


in  the  case  of  straight  lines,  it  is  usual  to  show  only  a  limited 
portion  of  a  cylindrical  surface,  but  it  really  extends  in- 
definitely in  both  directions. 

A  cylinder  is  a  solid  bounded  by  a  cylindrical  surface 
and  two  parallel  plane  surfaces  called  the  bases. 

The  altitude  of  a  cylinder  is  the  perpendicular  distance  be- 
tween the  planes  of  its  bases. 

A  right  cylinder  is  one  whose  straight  line  generators  are 
perpendicular  to  the  bases. 

29.  Cones.  A  conical  surface  is  the  surface  generated  by 
a  moving  straight  line  which  constantly  touches  a  fixed 
curve  and  passes  through  a  fixed  point  not  in  the  plane  of 
the  curve. 

A  cone  is  a  solid  bounded  by  a  conical  surface  and  a  plane 
which  cuts  all  positions  of  the  generating  line.  This  plane 
surface  is  called  the  base  of  the  cone. 


II,  §  29]        COMPUTATION  —  MEASUREMENT 


35 


The  altitude  of  a  cone  is  the  perpendicular  distance  from 
the  vertex  to  the  plane  of  the  base. 

A  right  circular  cone  is  one  whose  base  is  a  circle  and 
whose  altitude  meets  the  base  at  the  center  of  the  circle. 


Right  Circular  Cone 
Fig.  39 

The  slant  height  of  a  right  circular  cone  is  the  distance  along 
a  generator  from  the  vertex  to  the  circle. 

A  frustum  of  a  cone  is  the  portion  of  a  cone  included 
between  the  base  and  a  plane  parallel  to  the  base.  The 
frustum  thus  has  two  bases.  The  altitude  of  the  frustum 
is  the  perpendicular  distance  between  these  bases. 


Frustum  op  a  Cone 
Fig.  40 


The  lateral  surface  of  a  frustum  of  a  cone  is  the  portion  of 
its  lateral  surface  included  between  the  planes  of  its  bases. 

The  slant  height  of  a  frustum  of  a  right  cone  is  the  dis- 
tance between  the  bases  measured  along  a  generator. 


36 


MATHEMATICS 


[11,  §  30 


30.   The  sphere.    A  sphere  is  a  solid  bounded  by  a  surface 
all  points  of  which  are  equally  distant  from  a  point  within 


w- 

P 

^.1 

i 

D 

\ 

%b 

p' 

Sphere 
Fig.  41 


called  the  center.     A  sphere  may  be  generated  by  the  revo- 
lution of  a  semicircle  about  its  diameter  as  an  axis. 

31.   Surface  and  Volume  of  Solids. 

1.  Prism:  Volume=  area  of  baseX altitude, 

=  area    of    right    sectionXlateral 
edge. 
Lateral  Area  of  a  Right  Prism 

=  perimeter  of  baseX  altitude. 

2.  Cube:  Volume  =  s^  where  s  =  the  length  of  a  side. 

Entire  Surface  =  Qs^. 

3.  Cylinder:         Volume=  area  of  baseX altitude 

=  area     of     right     sectionX  slant 
height. 
Lateral  Area  of  a  Right  Cylinder 

=  perimeter  of  its  baseX altitude. 

4.  Pyramid:  Volume  =  one  third   of    area    of    baseX 

altitude. 
Lateral  Area  of  a  Regular  Pyramid 

=  half  perimeter    of    baseX slant 
height. 


II,  §  31]        COMPUTATION  —  MEASUREMENT 


37 


5.  Cone:  Volume  =  one  third  of  area  of  baseXalti- 

tude. 
Lateral  Area  of  Right  Circular  Cone 

=  half  perimeter  of  baseXslant 
height. 

6.  Frustum  of  a  Pyramid  or  of  a  Cone: 

Volume  =  ~(b-]-B-^y/bB) 

o 

where 

a  =  altitude  of  frustum, 
b  =  area  of  upper  base, 
B  =  area  of  lower  base. 

Lateral  Area  of  a  Frustum  of  a  Regular 
Pyramid  or  of  a  Right  Circular  Cone 

=  slant  heightXl/2  the  sum  of  the 

perimeters  of  the  bases 
=  slant   heightXthe  perimeter  of 
the  mid-section. 

7.  The  Sphere:    Volume  =  4:/Sirr^ 

where  r  =  the  radius. 
Surface  =  ATrr^, 

=  the  area  of  four  great  circles. 

8.  A  Barrel  :       It  may  be  considered  as  a  double  frustum 

of  a  cone  and  its  volume  and  surface 
approximated  accordingly.  But  for  cir- 
cular staves  a  closer  approximation  is  the 
following : 

Volume  =  -^  7r/i(2  D^+d^), 

in  which  i)  =  bung  diameter,  d  =  head 
diameter,  and  /i  =  height. 


Fig.  42.  —  Barrel 


38  MATHEMATICS  [II,  §  32 

32.  Similar  Figures.  In  general,  two  figures  are  similar 
if  they  have  the  same  shape. 

Two  polygons  are  similar  if  they  are  mutually  equiangular 
and  if  their  pairs  of  corresponding  sides  are  proportional* 

Similar  polyhedrons  are  those  that  have  the  same  number 
of  faces,  respectively  similar  and  similarly  placed  and  have 
their  corresponding  polyhedral  angles  f  equal. 

Any  two  similar  figures  in  a  plane  or  in  space  can  be  placed 
in  "  perspective,''  i.e.  so  that  straight  lines  joining  correspond- 
ing points  of  the  two  figures  will  pass  through  a  common 


Fig.  43.  —  Perspective 

point,  Fig.  43.  Thus  of  two  similar  figures  one  is  merely  an 
enlargement  of  the  other.  A  scale  drawing,  §  3,  and  the 
surface  it  represents  are  thus  similar. 

If  each  length  in  one  figure  is  k  times  the  corresponding 
length  in  the  other  figure,  then  each  area  in  the  first  figure  is 
fc2  times  the  corresponding  area  in  the  second.  Also  each 
volume  in  the  first  figure  is  k^  times  the  corresponding  volume 
in  the  second. 

This  may  also  be  expressed  as  follows  : 

Similar  surfaces  are  to  each  other  as  the  squares  of  their  like 
dimensions. 


*  To  say  that  corresponding  sides  are  proportional  means  that  each 
length  in  one  figure  is  a  constant,  k,  times  the  corresponding  length  in  the 
other  figure. 

t  The  opening  of  three  or  more  planes  which  meet  at  a  common  point  is 
called  a  polyhedral  angle. 


II,  §  32]        COMPUTATION  —  MEASUREMENT  39 

Similar  solids  are  to  each  other  as  the  cubes  of  their  like 
dimensions. 

Example.  Two  oranges  have  diameters  of  2"  and  3".  Compare 
their  surfaces  and  their  volumes,  respectively. 

Their  volumes  are  to  each  other  as  2^  is  to  3^,  i.e.  as  8  to  27.  Their 
surfaces  as  2^  to  3^,  i.e.  as  4  to  9. 

EXERCISES 

1.  Find  the  capacity  in  gallons  of  a  bucket  12"  deep  whose  bottom 
and  top  diameters  are  8"  and  10".  Find  the  whole  surface  of  this 
bucket. 

2.  Find  the  diagonal  of  a  barn  16'  X  30'  X  40'. 

3.  What  is  the  side  of  a  cube  whose  diagonal  is  2"? 

4.  Find  the  volume  in  cubic  feet  of  a  conical  haystack  whose  cir- 
cumference at  the  base  is  55  ft.  and  whose  slant  height  is  20  ft. 

5.  Find  the  number  of  gallons  of  vinegar  in  a  barrel  whose  head 
diameter  is  18  in.,  bung  circumference  70  in.,  and  height  35  in. 

6.  Find  the  volume  of  a  regular  pyramid  with  a  square  base  4  in. 
on  each  side,  if  the  altitude  of  the  pyramid  is  12  in.  Find  also  the 
lateral  surface  of  this  pyramid. 

7.  How  much  concrete  is  there  in  a  circular  silo  whose  walls  are 
8  in.  thick,  12  ft.  outside  diameter  and  30  ft.  high? 

8.  How  much  water  can  be  put  in  a  round  tank  8  ft.  deep,  if  the 
inside  diameters  are  4  ft.  at  the  bottom,  and  5  ft.  at  the  top? 

9.  The  circumference  of  a  sphere  is  1  ft.  2  in.  Find  its  area  and 
its  volume. 

10.  A  farmer  has  two  strings  of  drain  tile  3  in.  and  4  in.  in  diameter, 
respectively.  He  wishes  to  combine  them  into  one  equivalent  tile. 
What  should  be  its  diameter  ?     Does  a  tile  run  full  of  water  ? 

11.  What  would  be  the  size  of  a  waterspout  to  be  equivalent  to 
three  spouts  2  in.,  4  in.,  and  5  in.,  in  diameter? 

12.  How  many  2-in.  flues  of  an  engine  will  equal  in  sectional  area 
a  smokestack  22  in.  in  diameter? 

13.  A  farmer  has  a  triangular  field  whose  base  is  20  rods  and  whose 
sides  are  16  rods  and  12  rods.     Where,  along  the  side  16  rods  long, 


40  MATHEMATICS  [II,  §  32 

should  he  run  a  fence  parallel  to  the  base  in  order  to  divide  the  field 
into  two  equal  parts  ?  If  the  fence  is  run  parallel  to  the  base  and  starts 
at  the  middle  of  the  side  16  rods  long,  how  does  it  divide  the  field? 

14.  In  a  trapezoid  the  parallel  sides  8  ft.  and  12  ft.  long  are  16  ft. 
apart.  Find  the  altitude  of  the  triangle  formed  by  extending  the 
non-parallel  sides  of  this  trapezoid.    What  is  the  area  of  the  trapezoid? 

15.  In  the  previous  exercise,  where  should  a  line  be  run  parallel 
to  the  parallel  sides  to  divide  the  trapezoidal  field  into  two  equal  parts? 

Ans.     7.2  rd.  from  the  lower  base. 

16.  Grindstones  of  Huron  sandstone  will  safely  stand  a  surface 
speed  of  3600  ft.  per  minute.  How  many  revolutions  per  minute 
(R.  P.  M.)  may  a  3Ht.  stone  be  run?     A  2-ft.  stone?     A  6-ft.  stone? 

17.  A  hollow,  spherical  steel  shell  is  1  in.  thick  and  its  outside 
diameter  is  10  in.  The  specific  gravity  of  steel  is  7.8,  and  a  cubic  foot 
of  water  weighs  1000  oz.     Find  the  weight  of  the  shell. 

18.  The  side  of  an  equilateral  triangle  is  12  ft.  Find  its  altitude  and 
the  area. 

If  the  area  of  an  equilateral  triangle  is  400  sq.  ft.,  find  the  side. 
Make  a  formula  for  area  in  terms  of  the  side,  x,  say. 

19.  How  much  tin  is  there  in  a  fimnel  whose  diameters  are  1^  in.  and 

8  in.  and  whose  height  is  12  in.?     What  is  the  volume  of  this  funnel? 

20.  Find  the  volume  of  a  regular  hexagonal  pyramid  20  ft.  high 
if  each  side  of  the  hexagon  is  4  ft.  long.  Find  also  the  lateral  surface 
of  this  pyramid. 

21.  How  many  cubic  yards  must  be  excavated  in  digging  a  ditch 
200  rods  long,  12  in.  wide  at  the  bottom,  6  ft.  wide  at  the  top,  and  a 
yard  and  a  haK  deep  ? 

How  much  water  would  be  discharged  by  such  a  ditch  in  2  hr.  time 
if  it  flows  half-full  at  the  rate  of  1  ft.  per  second? 

22.  A  chimney  is  to  be  constructed  in  the  form  of  a  frustum  of  a 
circular  pyramid.  The  height  is  to  be  160  ft.,  with  outside  diameters 
7  ft.  and  10  ft.  The  flue  is  to  be  6  ft.  in  diameter  throughout  the  entire 
height.  How  many  2x4x8  in.  bricks  will  its  construction  require, 
allowing  10%  for  mortar? 

23.  Three  oil  cans  are  of  the  same  height.     Two  of  them  are  each 

9  in.  in  diameter,  and  the  other  one  is  13  in.  in  diameter.  Compare 
the  capacity  of  the  large  one  with  that  of  the  other  two. 


II,  §  32]       COMPUTATION  — MEASUREMENT  41 

24.  A  new  grindstone  is  30  in.  in  diameter  and  has  a  3-in.  face. 
What  has  it  lost  in  weight  when  it  is  worn  down  to  26  in.  in  diameter, 
if  sandstone  weighs  2.42  times  as  much  as  an  equal  volume  of  water, 
and  water  weighs  62.5  lb.  per  cubic  foot? 

25.  The  frustum  of  a  pyramid  is  8  feet  high,  and  its  bases  are  equi- 
lateral triangles  whose  sides  are  each  3  feet  and  4  feet,  respectively. 
Find  its  voliune. 

26.  In  the  previous  exercise,  find  the  volume  of  the  entire  pyramid. 

27.  The  height  of  a  certain  frustum  of  a  right  cone  is  §  the  height 
of  the  entire  cone.     Compare  the  volumes  of  the  frustum  and  the  cone. 

28.  A  lateral  edge  of  a  pyramid  is  6  feet.  At  what  distance  from 
the  vertex  should  this  edge  be  cut  by  a  plane  parallel  to  the  base  to  divide 
the  pyramid  into  two  parts  which  are  to  each  other  as  1  to  1?  As  1 
to  2?     As  3  to  4? 

29.  Find  the  volume  of  a  regular  tetrahedron  whose  edge  measures 
3  inches. 

[Hint.     The  tetrahedron  is  a  triangular  pyramid.] 

30.  A  wheat  bin  measures  4  ft.  by  6  ft.  by  10  ft.  What  are  the 
dimensions  of  a  similar  bin  that  holds  twice  as  much? 

31.  Find  the  area  of  the  solid  generated  by  an  equilateral  triangle 
that  revolves  about  one  of  its  sides  if  the  length  of  the  side  is  6  inches. 

32.  If  0.203  of  a  gram  of  gold  (specific  gravity,  19.32)  is  plated  over 
a  spherical  dome  whose  height  is  3  cm.,  and  the  radius  of  whose  base  is 
6  cm.,  what  is  the  thickness  of  the  gold? 

[Hint.  1  cc.  of  gold  weighs  19.32  grams.  Therefore  .203^19.32 
is  the  volume  of  the  gold.  This  divided  by  the  superficial  area  is  the 
thickness.] 

33.  A  steel  sphere  whose  radius  is  .95  cm.  weighs  28.25  grams. 
What  is  its  density?  Ans.     7.866 

34.  Funnels  are  best  made  with  an  angle  of  exactly  60°.  If  such 
a  funnel  measures  8  cm.  across  the  top,  what  size  filter  paper  will  fit 
it  flush  with  the  edge? 

35.  A  rectangular  block  of  wood  7.5  by  7.46  by  3.8  cm.  weighs 
152.7  grams.  What  is  its  specific  gravity  if  1  cc.  of  water  weighs  1 
gram? 

If  this  block  is  floated  in  water,  to  what  depth  will  it  sink? 


42 


MATHEMATICS 


[II,  §  33 


33.  Silos.  A  silo  is  a  receptacle  for  preserving  green 
feed,  and  silage  is  the  material  preserved.  A  silo  is  usually 
a  cylindrical  tank  much  higher 
than  wide.  Its  size  depends 
upon  the  number  of  animals, 
usually  cattle,  to  be  fed.  The 
diameter  of  the  silo  must  be  of 
such  size  as  to  insure  that  the 
proper  depth  will  be  removed 
daily.  Silage  of  all  kinds  de- 
teriorates unless  it  be  fed  regu- 
larly, evenly,  and  at  a  rate  of  not 
less  than  two  inches  depth  daily. 
Removing  five  or  six  inches  daily 
insures  but  little  waste  of  feed. 
Experience  has  shown  that  the 
most  satisfactory  results  are  ob- 
tained by  providing  a  horizontal  feeding  surface  of  about  5 
square  feet  for  each  cow.     See  Ex.  9  below. 

To  insure  proper  settling  of  the  silage,  excluding  in  this 
way  the  bacteria  that  cause  decay,  the  height  of  the  silo  should 
seldom  he  less  than  30  feet. 

On  an  average  one  ton  of  silage  occupies  50  cubic  feet. 

Thus,  the  diameter  of  a  silo  is  controlled  largely  by  the  num- 
ber of  cattle  to  be  fed,  while  the  height  is  gauged  by  the  quantity 
to  be  fed. 

One  cubic  foot  of  silage  per  head  is  a  widely  used  daily 
ration. 


FiQ.  44.  —  A  Cement  Silo 


EXERCISES 

1.  What  should  be  the  height  of  a  round  silo  for  a  herd  of  25  cows,  if 
each  one  is  to  be  fed  40  lb.  daily  for  180  days? 

Solution  :  For  25  cows  a  horizontal  feeding  surface  of  5  times 
25  =  125  square  feet  is  required.     This  is  the  area  of  a  circle  12.6  feet 


II,  §  33]        COMPUTATION  — MEASUREMENT  43 

in  diameter.     But  to  feed  25  cows,  each  40  lb.  daily  for  180  days, 
requires 

25  X  40  X  180  lb.  =  18,000  lb.  =  90  tons. 

Since  1  ton  occupies  50  cu.  ft.,  90  tons  will  require  4500  cu.  ft.  of 
space.  Therefore  the  height  is  the  capacity,  4500  cu.  ft.,  divided  by 
the  cross-sectional  area,  125  sq.  ft.,  i.e.  36  ft. 

2.  What  should  be  the  size  of  a  round  silo  to  feed  30  cows  40  lb. 
each  daily  for  150  days,  39  lb.  each  daily  for  100  days,  and  15  lb.  each 
daily  for  75  days  ? 

3.  What  should  be  the  size  of  a  round  silo  for  a  herd  of  20  cows,  the 
ration  being  40  lb.  each  daily  for  100  days  and  30  lb.  each  daily  for 
90  days? 

4.  Fattening  steers  are  fed  about  25  lb.  each  daily  and  calves  15  lb. 
each  daily.  How  large  a  silo  should  a  farmer  construct  to  accommodate 
in  this  way  40  steers  for  150  days  and  20  calves  for  100  days,  estimating 
40  lb.  as  the  daily  ration  for  an  animal  that  requires  5  sq.  ft.  of  hori- 
zontal feeding  surface. 

5.  What  is  the  capacity  of  a  round  silo  30  feet  high  and  15  feet  in 
diameter?  How  many  cows  will  it  maintain?  How  many  pounds 
daily  can  each  be  fed  from  this  silo  for  180  days? 

6.  What  depth  is  fed  daily  in  feeding  25  cows  each  40  pounds  per 
day  ?  If  a  smaller  daily  ration  is  contemplated,  how  should  the  above 
rule  for  size  of  a  silo  be  changed,  in  order  to  insure  proper  feeding  depth 
daily? 

7.  For  a  herd  of  15  cows,  each  one  fed  30  pounds  daily  for  175  days, 
what  should  be  the  minimum  size  of  the  silo? 

8.  A  silo  is  16  feet  in  diameter  and  36  feet  high.  What  is  the  least 
number  of  cows  that  must  be  kept  to  prevent  the  silage  from  spoiling, 
if  each  cow  is  fed  40  pounds  daily  ? 

9.  Explain  the  provision  of  5  sq.  ft.  of  horizontal  feeding  surface 
for  each  cow  on  the  basis  of  a  moderate  daily  feed  of  333  lb,,  an  average 
of  50  cu.  ft.  of  silage  per  ton,  and  a  minimum  of  2"  depth  removed  daily. 


CHAPTER  III 

INDIRECT  MEASUREMENT  — TRIGONOMETRY  — 
SURVEYING 

34.  Proportion  and  Indirect  Measurement.  A  propor- 
tion is  an  equality  of  ratios,*  as 

2  =  1.,  M  =  12  Ib.^  4hr.  ^  20  mi. 

3  12'  $4      16  lb.'  7  hr.      35  mi. 

The  proportion  -  =  -  may  also  be  written  in  the  form 
6     a 

a:h  =  c:  d. 

There  are  four  terms  in  a  proportion.     If  any  three  of  them 
are  known  the  fourth  may  be  found.     Thus,  if  we  have 

a     c 
multiplying  each  side  by  a, 

xv.     oh 

X  =  — 

c 

In  stating  a  proportion  where  one  of  the  terms  is  un- 
known it  is  convenient  to  put  the  unknown  term  first  as  just 
shown. 

To  estimate  the  height  of  a  ham,  a  farmer  has  an  assistant 


*  A  ratio  is  the  relation  of  one  number  to  another  of  the  same  kind  ex- 
pressed by  their  quotient.  The  ratio  of  a  to  6  is  a/b.  Every  fraction 
expresses  the  ratio  of  its  numerator  to  its  denominator  and  every  integer 
expresses  the  ratio  of  itself  to  unity. 

44 


Ill,  §  34] 


INDIRECT   MEASUREMENT 


45 


hold  a  staff  NB  vertically,  Fig.  45.     As  he  sights  along  AE 
to  the  top  of  the  barn,  the  assistant  marks  the  point  B  on 


M  N 

Fig.  45.  —  Indirect  Measurement 

the  staff.     When  a  =  BC,  b  =  AC,  and  AD  are  measured,  the 
distance  x  =  DE  may  be  computed  from 

The  height  of  the  barn  is  then  x  plus  the  height  of  the  ob- 
server. 

The  principle  used   here  forms  the  basis  of  nearly  all 
practical  methods  of  indirect  measurement. 


EXERCISES 

1.    Find  the  height  of  a  barn,  when,  as  in  Fig.  45, 
AD  =  75',  and  the  observer  is  5'  tall. 


2',   6  =  5', 


2.  A  post  12'  high  casts  a  shadow  30'  long.  How  high  is  a  tree 
whose  shadow  at  the  same  time  is  160'  ? 

State  the  proposition  from  elementary  geometry  which  is  used  in 
the  solution  of  this  problem. 

3.  At  a  distance  of  62'  from  a  building  is  a  vertical  post  10'  high. 
By  standing  back  of  the  post  6'  and  sighting  over  the  end  of  a  4'  vertical 
stick,  the  top  of  the  post  and  the  top  of  the  building  are  in  line.  How 
high  is  the  building? 

4.  A  carpenter's  square  is  16"  X  24",  the  blade  is  1|"  wide,  and  the 
arm  is  2"  wide.  A  man  places  the  blade  vertically  on  top  of  a  post 
5'  high  and  40'  from  a  building.  How  high  is  the  building,  if  the  line 
of  sight  over  the  arm  and  blade  just  intersects  the  top  of  the  building? 


46 


MATHEMATICS 


[in,  §  34 


Fig.  46 


5.  The  line  of  sight  between  two  objects  A  and  B,  Fig.  46,  is  ob- 
structed by  a  building.  The  distance  AB  may  now  be  estimated  by 
measuring  the  distances  AO  and  BO  from  some 
convenient  point  0,  laying  off  OC  =  OA  and 
OD  =  OB,  and  finding  DC.     Show  why. 

6.  The  distance  AB,  Fig.  47,  across  a  swamp 
or  river  may  be  estimated  by  use  of  a  vertical 
staff  to  which  is  attached  an  arm  CD  at  right 
angles  to  the  staff  and  movable  up  and  down  along  it.     By  sighting 
along  EB  and  knowing  AE,  DE,  DC,  the  distance  AJ5  is  found. 

If  AE  =  7',  DC  =  S',  and  ED^^",  find 
the  distance  across  the  swamp. 

Moreover,  an  accessible  point  B'  may  be 
located  so  that  AB'  =  AB  by  simply  re- 
volving the  arm  DC  around  the  staff  and 
sighting  to  B'  in  the  line  EB'. 

This  device  was  used  in  the  early  days, 
and  formed  the  basis  of  the  surveying  instruments  used  in  those  times. 

7.  A  boy  holds  a  pencil  6"  long  2'  from  his  eye  so  that  it  covers  a 
chimney  200'  distant.  What  is  the  height  of  the  chimney?  Draw  a 
figure,  and  explain  the  process. 

8.  Draw  a  line  segment.  Divide  it  into  three  equal  parts.  Into 
five  equal  parts. 

9.  The  following  figure  represents  a  diagonal  scale  used  by  drafts- 
men.    The  distance  from  0  to  10  is  1". 


.^' 


Fig.  47 


10  987654321 


Fig.  48.  —  Diagonal  Scale 


Explain  the  principle  involved  in  the  use  of  such  a  scale. 
Show  how  to  set  the  dividers  to  the  following  distances :   .27' 
1.89",  1.04". 


.73' 


Ill,  §  36] 


INDIRECT  MEASUREMENT 


47 


35.  Angles  and  the  Sides  of  a  Right  Triangle.  A  right 
triangle  is  formed  by  dropping  a  perpendicular  line  from 
any  point  B  in  the  terminal  of  an  angle,  upon  the  initial  line 
at  C,  Fig.  49.     Thus  a  =  BC  is  the  side  opposite  the  angle  A, 


Fig.  49 

h  =  AC  is  the  side  adjacent  to  the  angle  A,  and  AB  is  the 
hypotenuse  of  the  right  triangle. 

Now  as  the  angle  A  changes,  the  sides  of  this  right  triangle 
will  change.  The  amount  of  turn,  or  the  size  of  the  angle,  thus 
depends  upon,  or  is  related  to,  the  sides  of  the  right  triangle, 
as  shown  in  the  next  few  paragraphs. 

36.  Use  of  Instruments.  Ratios.  Angles.  In  the  study 
of  indirect  measurement,  §  34,  we  saw  that  the  ratios  of 
sides  and  hypotenuse  of  right  triangles  play  an  important  part 
in  solving  problems  involving  triangles.  This  is  particularly 
true  in  finding  heights  and  distances  and  in  surveying. 

Where  more  exact  methods  are  desired,  a  surveyor's 
transit,  or  a  sextant,  or  some  similar  instrument,  is  used  to 
find  the  ratio  of  the  sides, 
instead  of  using  a  staff  and 
sighting  as  was  done  in 
§  34.  For  example,  sup- 
pose we  wish  to  know  the 
height  of  a  tree  CB,  as  in 
Fig.  50.  The  angle  A  could  ^'''  ^^ 

be  measured  directly  by  means  of  one  of  these  instruments. 
The  ratio  desired,  a/h  in  this  case,  is  called  the  tangent;  it 


48  MATHEMATICS  [III,  §  36 

can  be  found  from  tables  which  are  specially  prepared  for 
this  purpose.  Suppose  the  angle  at  A  is  found  to  be  25°  20'. 
Then  from  such  a  table  (in  the  column  for  tangents,  Table 
IX,  Appendix),  we  find  opposite  25°  20'  the  value  .4734. 
This  means  that  a/b  =  .4734.  If  h  is  measured  and  is  found 
to  be  100',  then  a  =  (.4734)  100' =  47.34',  the  height  required. 
Moreover,  if  we  know  the  sides  and  therefore  their  ratio, 
such  a  table  gives  us  the  angle  which  corresponds  to  that  ratio. 
For  example,  if  the  ratio  a/b  =  .5467,  we  find  from  the  table 
of  tangents  that  this  is  the  ratio  for  28°  40' ;  hence  A  =  28°  40'. 

37.  Trigonometric  Ratios.  For  convenience  these  ratios 
in  the  right  triangle  have  been  given  special  names. 

Draw  a  triangle  ABC  right-angled  at  C,  Fig.  51,  naming, 
as  is  usual,  the  angles  with  the  capitals  and  the  sides  opposite 
the  angles  by  the  corresponding  small  letters. 


FiQ.  51 

Definitions. 

Tangent  A  is  the  ratio  of  the  side  opposite  angle  A  to  the 
side  adjacent  to  A,  or  a/b  in  Fig.  51. 

Sine  A  is  the  ratio  of  the  opposite  side  to  the  hypotenuse, 
or  a/c  in  Fig.  51. 

Cosine  A  is  the  ratio  of  the  adjacent  side  to  the  hypotenuse^ 
or  h/c  in  Fig.  51. 

Abbreviations  for  these  expressions  are  : 

tanA=-,  sinA=-,  cosA=—' 
b  c  c 

These  are  called  trigonometric  ratios. 

In  this  same  triangle  write  the  similar  ratios  for  tan  B,  sin  B,  cos  B. 


Ill,  §  37] 


INDIRECT  MEASUREMENT 


49 


EXERCISES 

Find  the  value  of  the  unknown  term  in  each  of  the  following  exercises, 
expressing  this  value  decimally  also. 


1.   5 


4.   23  = 


2.     9 


3.   27  = 


,7 


5.    .4759  = 


35.6 


6.  If  tan  32°  =  -^,  find  a. 
77. o 


Ans.    a  =  (77.3)  tan  32°. 


18 


7.   Solve,  as  in  Ex.  6,  (a)  sin  A  =  —  ,  for  c. 

c 

(b)  tan  25°  =^,for6. 

0 

/    X        cos  20°  1  Q      r 

(c)    =  13,  for  a. 


8.  For  a  right  triangle  ABC,  with  sides  a,  b,  c,  as  in  Fig.  51,  fill  in 
the  omitted  entries  in  the  following  table  from  the  geometric  properties 
of  these  special  triangles.     Express  the  ratios  decimally. 


Ex. 

a 

6 

c 

sin  A 

cos  A 

tan  A 

angle  A 

(a) 

3 

6 

30° 

(b) 

27 

27 

45° 

(c) 

15 

.5 

60° 

id) 

17 

30° 

9.   If  o  =  7,  6  =  24,  find  sin  A,  cos  A,  tan  A. 

tan  B. 


Also  find  sin  B,  cos  B, 


10.  A  rafter  rises  5'  in  a  horizontal  run  of  12'.  What  is  the  length 
of  the  rafter?  What  are  the  sine  and  the  cosine  and  the  tangent  of 
the  angle  which  this  rafter  makes  with  the  horizontal?     See  Fig.  11. 

11.  The  side  of  an  equilateral  triangle  is  10.4'.  Find  its  altitude 
and  its  area.     If  the  side  is  s,  find  its  altitude  and  its  area  in  terms  of  s. 

12.  What  is  the  length  of  the  side  of  an  equilateral  triangle  which 
contains  1  acre? 

13.  Is  the  sine  a  proper  or  an  improper  fraction?  Why?  Is  the 
cosine  greater  than  unity  or  less  than  unity?  How  large  may  the 
tangent  become? 


50  MATHEMATICS  [III,  §  38 

38.  Tables.  The  values  of  the  trigonometric  ratios  for 
angles  differing  by  10'  have  been  computed  to  four  decimal 
places  and  recorded  in  Table  IX  in  the  Appendix  of  this 
book.  They  are  arranged  with  the  angles  from  0°  to  45° 
reading  down  the  left  side  of  the  page  and  the  name  of  the 
desired  functions  of  these  angles  at  the  top  of  the  page. 
Angles  from  45°  to  90°  are  read  up  the  right  side  of  the  page 
and  the  name  of  the  functions  at  the  bottom  of  the  page. 

By  means  of  computed  corrections  these  tables  may  be 
read  to  the  nearest  minute.  These  corrections  are  made 
according  to  the  rule  that  relatively  small  changes  in  the  angle 
are  proportional  to  the  corresponding  changes  in  the  value. 

The  use  of  the  tables  is  shown  by  the  following  examples.* 

Example  1.  Find  sin  37°  20'.  On  the  proper  page  of  the  table 
find  37°  20'  in  the  left-hand  column  marked  Degrees  at  the  top.  Directly 
opposite  37°  20'  in  the  column  marked  Sine-Value  at  the  top,  we  find 
.6065      Thus  sin  37°  20' =  .6065     Similarly,  sin  37°  40' =  .6111 

Example  2.     Find  sin  37°  24'.     Find,  in  this  case, 
sin  37°  20' =  .6065 
sin  37°  30' =  .6088 
Thus  10'  change  in  the  angle  =  a  change  of  23  in  the  value  of  the  sine  and 
therefore  4'  change  in  the  angle  =  a  change  of  ^  of  23  =  9  in  the  sine,  i.e. 
sin  37°  20' =  .6065 
correction  for  4'  =        9 
Hence  sin  37°  24' =  .6074 

Example  3.     Find  tan  37°  47'.     Here 
tan  37°  40' =  .7720 
tan  37°  50'  =  .7766 
Therefore,  for  10',  the  correction  is  46. 

and  for  7',  the  correction  is  j^  of  46  =32. 

Thus  tan  37°  40' =  .7720 

correction  for  7'  =      32 
Therefore  tan  37°  47' =  .7752 


*  For  more  extensive  tables,  giving  values  for  every  minute  of  arc  to  five 
decimal  places,  see,  e.g.,  The  Macmillan  Tables. 


Ill,  §  38] 


INDIRECT  MEASUREMENT 


51 


Example  4.     Find  cos  19°  12'.     We  have 
cos  19°  10'  =  .9446 
cos  19°  20' =  .9436 


and 


correction  for  10'  =  10 
correction  for  2' =  2. 


This  correction  is  subtracted  in  the  case  of  the  cosine,  since  the  valiie 
of  the  cosine  becomes  smaller  as  the  angle  increases. 
Thus  cos  19°  12' =  .9444 

Example  5.     If  sin  A  =  .6870,  find  A.     Not  finding  precisely  this 
number  in  the  table,  take  the  one  next  nearest  below  it.     Thus 
.6862  =  sin43°20' 
■6870  =  sin  A 

8  =  correction  to  be  found  for  the  angle. 


But 

.6884  =  sin43°30'. 

Therefore 

22  =  correction  for  10' 

and 

8  =  correction  for  -^  of  10' =  4'. 

Hence 

.6870  =  sin  43°  20'-f4'  =  sin  43°  24'. 

Example  6. 

If  tan  A  =  1.6795,  find  A. 

1.6753=  tan  59°  10', 

1.6795  =  tan  A, 

1.6864  =  tan59°20'. 

Therefore 

Ill  =  correction  for  10' 

and 

42  =  correction  for  ^^  of  10'  =4', 

whence 

1.6795  =  tan  59°  24',  i.e.     A  =  59°  24'. 

EXERCISES 

With  the  table  just  explained,  read  and  tabulate  the  sine,  cosine, 
and  tangent  of  each  angle  in  the  following  table : 


Ex. 

Angle 

sin 

cos 

tan 

Ex. 

Angle 

sin 

cos 

tan 

1 

32°  34' 

7 

71°  9' 

2 

27°  53' 

8 

27°  7' 

3 

41°  22' 

9 

81°  29' 

4 

11°  19' 

10 

68°  41' 

5 

56°  48' 

11 

84°  16' 

6 

61°  17' 

12 

64°  54' 

52 


MATHEMATICS 


[HI,  §  39 


39.  Construction  of  Angles.  An  angle  may  be  laid  off 
directly  by  means  of  a  protractor.  However,  it  can  be 
constructed  more  accurately  by  means  of 
the  value  of  its  tangent  obtained  from  a 
table.  When  squared  paper  *  is  used  neither 
protractor  nor  compasses  are  necessary. 


Example  1.     Given,  tan  A  =  .75,  to  construct  A. 

Since  ■75  =  -^^'jj  =  ^,  lay  off  4  squares  horizontally, 

Fig,  52,  and  3  squares  vertically   and   draw  the 


B 

y 

^/ 

y^ 

a 

y 

h 

A 

^ 

Fig.  52 
hypotenuse  AB.     Then  angle  CAB  is  the  required  angle. 


Example  2.     Construct  K  if  sin  ii[  =  5/13. 

Here  the  hypotenuse  is  13  and  the  side  opposite  X  is  5.  The  third 
side  of  the  right  triangle  is  therefore  VlS^  — 5^  =  12  and  tan  K  =  5/12. 
The  construction  then  follows  as  just  shown. 

EXERCISES 

Draw  a  right  triangle  to  which  each  of  the  following  functions  belongs 
and  give  the  numerical  value  of  each  of  the  other  two  functions. 

1.  tanvl=j\.  3.   sin  5=^.  6.   sin/^  =  .3 

2.  cosA:  =  t\.  4.   tanfl'  =  2i.  6.   cosA=f. 
Construct  the  following  angles  from  the  value  of  their  tangents : 

7.  37°.  9.   54°  30'.  11.   84°. 

8.  20°.  10.   5°.  12.   60°. 

40.  Solution  of  Right  Triangles.  To  solve  a  right  triangle 
is  to  compute  the  unknown  parts.  Two  fundamentally 
different  cases  exist. 

I.  When  two  sides  are  given. 

II.  When  a  side  and  an  acute  angle  are  given. 

In  the  solution  of  problems,  here  as  elsewhere,  the  given  data 
should  be  drawn  carefully  to  a  convenient  scale.  This  in 
itself  is  an  approximate  solution. 

*  See  §  78.  This  is  merely  paper  ruled  into  small  squares  or  small 
rectangles.     It  can  be  purchased  at  most  students'  supply  stores. 


III.  §  40] 


INDIRECT  MEASUREMENT 


53 


EXERCISES 

Solve  each  of  the  following  right  triangles.     It  is  assumed  that  the 
triangle  is  lettered  as  in  Fig.  51  unless  special  lettering  is  given. 

1.   Given  c  =  12.5,  A  =  17"^  24'.     Find  a,  b,  B. 
Solution  :  B  =  90°  - 17°  24'  =  72°  36', 


hence 


-^=sinl7°24'  =  .2990; 

a  =  (12.5)  (.2990)  =3.74 


12.5 


=  cosl7°24'  =  .9542, 


whence  b  =  (12.5)  (.9542)  =  11.9 

2.   Given  b  =  140,  B  =  24°  12'.     Find  a,  c,  A. 
Solution  :  A  =90° -24°  12' =  65°  48' 

-^  =tan  65°  48' =  2.2251 
140 

a  =  (140)(2.2251)=311.5 

140 

c 

140 


whence 


sin  24°  12' =  .4099, 
=  341.5 


.4099 

3.  Given  c  =  340,  5  =  29°.  Ans. 

4.  Given  b  =  270,  A  =  54°.  Ans. 

5.  Given  6  =  42.4,  B  =  57°  46'.     Ans. 

6.  Given  a  =  3,  6  =  4. 


Fig.  54 
A=61°,  6  =  165,  a  =  298. 
B  =  36°,  a  =  372,  c  =  459. 
A  =32°  14',  a  =  26.7,  c  =  50.1 


Solution 


whence 


Check: 


tan  A=l  =  .75 
A  =36°  52'. 

^=sin36°52'  =  .6000, 
c 

a2+b2  =  c2,  i.e.  32+42  =  52. 


7.  Given  a  =  37,  6  =  121. 

8.  Given  c  =  5.4,  a  =  2.6. 
[Hint.     Use  sin  A  =a/c.] 

9.  Given  c= 63,  6  =  47. 


Ans.    A  =  17°,  5  =  73°,  c  =  127. 

Ans.     A=28°47',  5  =  61°  13',  6=4.7 

Ans.    A  =  41°  45',  B = 48°  15',  a  =  42. 


54  MATHEMATICS  [III,  §  40 

10.  Two  sides  of  a  triangle  are  37.4  and  29.8,  and  their  included 
angle  is  31°  47',  as  shown  in  the  figure.  Find  the  third  side,  a,  and  the 
angles  B  and  C  of  this  triangle  and  the 
altitude,  h,  on  AC. 

[Hint.  Find  p  =  AD  and  h  in  the  right 
triangle  ADB  from  p  =  (29.8)  cos  31°  47', 
etc.     Then  g  =  37.4 -p.] 

11.  Ifc  =  13.37,A  =  13°13',  finda,  fe,i5. 

12.  Given  a  =  12.78,  c  =  13.42  in  a  right  triangle.  Find  the  angles 
and  the  third  side. 

13.  A  tile  drain  has  a  fall  of  4"  to  the  rod.  What  angle  does  it 
make  with  the  horizontal  in  the  same  vertical  plane  with  the  drain  ? 

14.  What  is  the  length  of  the  rafters  for  a  house  28'  4"  wide  if 
they  have  a  rise  of  1  in  2  of  run,  and  project  16"  beyond  the  eaves? 
What  angle  does  each  make  with  the  horizontal  in  the  same  vertical 
plane  with  the  rafter? 

15.  Each  of  two  sides  of  an  isosceles  triangle  is  27  rods  long  and  the 
angle  at  the  vertex  is  47°  38'.     Find  the  base,  altitude,  and  area. 

[Hint.     Bisect  the  given  angle  and  solve  one  right  triangle.] 

16.  A  corner  post  is  anchored  to  a  stone  buried  in  the  ground  by  a  stay 
wire  which  makes  an  angle  of  25°  with  the  horizontal.  If  the  tension  in 
the  wire  is  200  pounds,  find  the  vertical  ^  ^ 
and  the  horizontal  pull  on  the  stone. 

17.  Two  sides  of  a  parallelogram  are 
12'  and  15',  and  the  included  angle  is    ^  ib  b  a 
30°  35'.     Find  the  two  diagonals.                                   ^^^'  ^^ 

[Hint.  Find  the  horizontal  and  the  vertical  projections  of  EC. 
Then  AC  follows  from  the  right  triangle  AEC] 

18.  Two  forces  of  40  pounds  and  50  pounds,  respectively,  act  on  a 
body  at  an  angle  of  40°  52'.  Find  the  resultant  of  these  forces  and  the 
angle  it  makes  with  either  force. 

[Hint.  Find,  as  in  the  previous  exercise,  the  diagonal  of  the  par- 
allelogram whose  sides  represent  to  scale  these  forces.] 

19.  The  draft  of  a  certain  plow  is  rated  at  296  lb.  What  power 
must  be  supplied  to  operate  it  if  the  traces  of  the  horses  pulling  it  make 
an  angle  of  11°  with  the  horizontal? 


Ill,  §  41] 


INDIRECT  MEASUREMENT 


55 


41.  Angles  of  Elevation,  of  Depression.  The  angle  which 
a  line  from  the  eye  to  an  object  makes  with  a  horizontal 
line  in  the  same  vertical  plane  is  called  an  angle  of  eleva- 
tion if  the  object  is  above  the  eye  of  the  observer  and  an 
angle  of  depression  if  the  object  is  below  the  eye  of  the 
observer,  Fig.  58. 


Fig.  58 

Example.  The  angle  of  elevation  of  the  top  of  a  barn  is  42°  40'  at 
a  point  102.4'  measured  horizontally  from  the  base.  Find  the  height 
of  the  barn. 

From  the  right  triangle,  we  have 


=  tan  42M0'  =  . 9217 


h'? 


102.4' 

Hence         ;i  =  (102.4)  (.9217) 

=  94.38',  the  height  required. 


EXERCISES 

1.  What  is  the  angle  of  elevation  of  the  sun  when  a  10'  pole  casts 
a  shadow  8'  long  ? 

2.  What  is  the  height  of  a  tree  if  the  angle  of  elevation  of  its  top 
is  37°  35'  measured  at  a  point  147.8'  horizontally  from  its  foot? 

3.  From  the  top  of  a  cliff  167'  high  the  angle  of  depression  of  a 
man  on  the  plain  is  27°  28'.  How  far  is  the  man  from  the  base  of  the 
cliff? 

4.  A  triangular  field  ABC  contains  one  acre.  If  Z  CAB  =  52°  8' 
and  Z  B  =  90°,  find  AB  and  BC. 

[Hint.  Let  AB  =  x  and  find  BC  in  terms  of  x  from  tan  CAB.  The 
area  in  terms  of  x  may  now  be  found,  etc.] 

5.  What  is  the  rise  in  one  third  of  a  mile  in  a  road  which  has  a 
uniform  grade  of  2°  30'? 


56  MATHEMATICS  [III,  §  41 

6.  A  tower  and  a  building  stand  on  a  level  plane.  At  a  window 
23'  from  the  ground,  the  angle  of  depression  of  the  base  of  the  tower 
is  13°  30'  and  the  angle  of  elevation  of  the  top  is  39°  47'.  Find  the 
height  of  the  tower, 

7.  Two  men  are  lifting  a  stone  by  means  of  ropes.  The  ropes  are 
in  the  same  vertical  plane.  What  is  the  weight  of  the  stone  if  one  man 
pulls  105  lb.  in  a  direction  48°  with  the  horizontal  and  the  other  one 
85  lb.  in  a  direction  67°  with  the  horizontal?  Ans.     156  lb. 

8.  A  hillside  farm  contains  52  acres,  actual  area.  It  is  located  on 
a  slope  of  16°.  How  much  more  is  this  than  the  horizontal  surface 
underneath  it  and  between  the  same  vertical  bounds? 

Ans.     2  acres. 

9.  At  a  point  A  the  angle  of  elevation  of  the  top  of  a  tree  was  32**. 
At  B,  160'  from  A,  the  elevation  of  the  tree  was  found  to  be  21°.  What 
is  the  height  of  the  tree  if  A,  B,  and  the 
tree  are  in  a  straight  line  and  in  the  same 
level  plane? 

[Hint.     Drop  a  perpendicular  AD  on 
BE,  Fig.  54.     Z  AED  is  equal  to  32° -21°    ^       ^^'  ^ 
=  11°.     Find  AD  from  the  right  triangle 

ABD.     When  AD  is  known,  compute  AE  from  the  triangle  AED. 
Then  find  EC] 

10,  Find  the  height  of  a  precipice  if  its  angles  of  elevation  at  two 
stations  in  a  horizontal  line  with  its  base  are  41°  37'  and  34°  13'  and  the 
distance  between  stations  is  145'. 

42.  Farm  Surveys.  Farms  are  surveyed  by  measuring 
distances  and  angles  along  the  boundary  of  the  farm  and 
fixing  the  corners  by  permanent  landmarks  such  as  boulders, 
iron  pins,  concrete  posts,  etc.  The  data  obtained  in  this 
way  are  then  plotted  to  scale. 

The  area  of  the  survey  is  usually  computed  by  drawing 
two  axes  —  north-south  and  east-west  lines  —  so  that  the 
scale  diagram  of  the  survey  lies  entirely  to  the  right  and 
above  these  lines  respectively.  Perpendiculars  are  drawn 
from  each  vertex  to  this  north-south  line,  forming  as  many 
trapezoids  {or  triangles)  as  the  survey  has  sides. 


Ill,  §  43]  INDIRECT  MEASUREMENT  57 

43.  Bearing  of  a  Line  or  Course.  Northing  (Southing), 
Easting  (Westing)  of  a  Course.  The  bearing  of  a  course 
means  its  direction  by  compass.  N.  40°  E.,  Fig.  61,  means 
the  direction  reached  after  sighting  north  and  then  swinging 
the  line  of  sight  40°  toward  the  east,  i.e.  angle  NOD  =  4:0°. 

When  going  around  a  plot  or  parcel  of  land  to  determine 
the  lengths  and  bearings  of  all  the  sides,  the  surveyor  bears 
north  or  south,  east  or  west  on  each 
course,   distances    respectively   equal 
to  the  projections  of  the  course  on 
these  north-south,  east-west  lines.   For 

example,  the  course  OD,  Fig.  61,  bears   ^ 

north  a  distance  OH,  called  the  north- 
ing (southing),  and  bears  east  a  dis- 
tance  HD,  called  its  easting  (west- 
ing).    For  a  closed  survey,  the  computation  should  show 

northings  =  southings, 
and 

eastings  =  westings. 

For  the  course  OD,  Fig.  61, 

northing  0H  =  OD  cos  40°, 
and 

easting  HD  =  OD  sin  Jfi°. 

In  general,  for  any  course  we  have, 
northing  (southing)  =  length  of  course  times  cosine  of  its  hearing, 
and 

easting  (westing)  =  length  of  course  times  sine  of  its  hearing. 

Northing  (or  southing)  of  a  course  is  the  difference  of 
latitudes  of  its  end  points,  easting  (westing)  is  the  difference 
of  longitudes  of  its  end  points. 


58 


MATHEMATICS 


[in,  §  43 


Example  1.     Plot  the  following  courses  and  compute  their  latitudes 

and  longitudes,  i.e.  their  northings  or  southings,  j^ 

eastings  or  westings.  b' 

A  line  is  run  from  a  point  A,  N.  35°  E.,   12 
chains  to  a  point  B ; 

from  B,  S.  35°  E.,  10  chains  to  a  point  C;  c' 

from  C,  S.  82°  36'  W.,  12.7  chains  to  the  place  ^ 

of  beginning.  ^ 


Fig.  62 


Then,  AB',  the  northing  on  AB  =  12  cos  35°  =  9.83. 
B'B,  the  easting  on  AB  =  12  sin  35°  =  6.88 


CC,  the  westing  on  CA  =  (12.7)  sin  82°  36'  =  12.59 
This  may  be  arranged  as  follows : 


Course 

Bearing 

Distance 

Latitudes 

Longitudes 

Northing 

Southing 

Easting 

Westing 

AB 
BC 
CA 

N.  35°  E. 
S.  35°  E. 
S.  82°  36'  W. 

12  ch. 
10  ch. 
12.7  ch. 

9.83 

8.19 
1.63 

6.88 

5.74 

12.59 

Example  2.  Plot  the  following  survey  to  scale  and  find  the  area 
of  the  inclosed  tract  by  projecting  the  courses  on  the  north-south  line 
through  the  vertex  farthest  west. 

Beginning  at  a  post  A,  thence  N.  3°  E.,  5.60 
ch.  to  a  stone  12"  long  and  5"  square,  thence  S. 
84°  E.,  7.50  ch.  to  a  post,  thence  S.  10°  30'  E.,    c' 
6.42  ch.  to  an  iron  pin,  thence  N.  80°  30'  W., 
9.00  ch.  to  the  place  of  beginning. 


The  scale  diagram  is  shown  in  Fig.  63.     The    d' -TrtrsJ^) 

computation  may  be  arranged  as  follows :  Fig.  63 

For  the  first  course  AB,  N.  3°  E.,  5.60  chains,  we  have 

latitude  north,  AB'  =  (5.60)  cos  3°  =  (5.60)  (.9986)  =  5.592, 
longitude  east,  B'B  =  (5.60)  sin  3°  =(5.60)  (.0523)=  .287 


Ill,  §  43] 


INDIRECT   MEASUREMENT 


59 


These  and  similar  results  for  the  other  courses  give  the  following 
tabulation. 


Bearing 

Distance 

COMPUTED 

BALANCED 

Course 

Latitudes 

Longitudes 

Latitudes 

Longitudes 

North- 
ing 

South- 
ing 

East- 

West- 
ing 

N 

S 

E 

W 

AB 
BC 
CD 
DA 

N3°E 
S84°E 
S10°30'E 
NSO'SO'W 

5.60  ch. 
7.50  ch. 
6.42  ch. 
9.00  ch. 

5.592 
1.485 

.784 
6.313 

.293 
7.443 
1.170 

8.877 

5.596 
1.491 

.779 
6.308 

.287 
7.435 
1.164 

8.886 

28.52 

7.077 

7.0j7 

7.077 

.02 

8.906 

8.877 

.029 

8.877 

7.087 

7.087 

8.886 

8.886 

Thus  the  error  in  latitudes  is  .02  ch. ;  in  longitudes  the  error  is  .029  ch. 
The  survey  is  balanced  by  distributing  these  errors  over  the  total  perimeter, 
28.52  ch.     Therefore, 


error  in  latitudes  for  1  chain  = 


.02 


error  in  longitudes  for  1  chain 


28.52 

.029 
28.52 


.0007; 


.001; 


and  error  in  AB,  5.60  ch.  =  (5.60)  (.0007)  =  .004  for  latitudes 
=  (5.60)  (.001)  =  .006  for  longitudes. 

Similar  results  may  be  calculated  and  tabulated  for  the  remaining 
courses  BC,  CD,  DA. 

Since  the  northings  are  too  small,  add  the  corrections;    similarly, 
subtract  them  from  the  southings.     Likewise,  balance  the  eastings  and 
westings.     This  gives  the  balanced  computation  in  the  diagram  above. 
The  Area  ABCD  =  C'B'BC+CDD'C'  -ADD'  -ABB' . 
C'B'BC  =  hC'B'(B'B-\-C'C)  =  h  (.779) (.287+7.722)      =  3.120 
CDD'C  =  h  C'D'iC'C+D'D)  =  h  (6.308) (7.722+8.886)  =52.382 

55.502 
ADD'  =  hAD''  D'D  =  \  (1.491)  (8.886)  =6.625 
ABB'  =  i  A^'  •  B'B  =  h  (5.596)  (.287)   =   .813 

7.438 
Hence,  Area  =  48.064  sq.  ch. 

=4.8064  A. 


60 


MATHEMATICS 


[III,  §  43 


EXERCISES 

1.  Find  the  number  of  acres  in  a  triangular  field  whose  sides  are 
14.6  ch.,  17.3  ch.,  and  19.7  ch. 

2.  Find  the  area  of  a  triangular  field  that  has  two  sides  16.6  ch.  and 
31.2  ch.,  and  the  included  angle  63°  25'. 

3.  Find  the  area  of  a  triangular  field  of  which  two  angles  are 
67°  33'  and  48°  28'  and  the  included  side  is  q 

20.4  ch. 

[Hint.  Find  the  third  angle.  Then  find 
the  altitude  /i  =  (20.4)  sin  48°  28';  also  find 
AK.  Since  angle  B  is  found,  KB  is  equal  to 
^-f-tan  B.  The  area  then  follows  from  the 
base  AB  and  the  altitude  /i.j 

4.  A  man  has  a  triangular  shaped  piece  of  land  ABC  as  shown  in 
the  accompanying  figure.  The  side  AB  bears  N.  59°  20'  E.,  27.30  rd. 
and  BC  bears  N.  46°  10'  W.,  26.80  rd. 

Find  the  length  and  the  direction  of  AC  and 
the  area  of  the  field  assuming  that  the  given 
measurements  are  correct. 

[Hint.  Compute  the  longitudes  EB  and  BK. 
Their  difference  is  CD.  Similarly  compute  DA. 
But  CD  and  DA  determine  the  direction  of  AC. 
The  area  ABC  =  BCDE+EBA-ACD.] 

Ans.     CA  bears  S.  7°  8'  W. 

Plot  the  following  surveys  to  scale,  balance  the  computation  for 
distribution  of  error  of  closure,  and  find  the  area. 

5.  Beginning  at  a  stone  in  the  center  of  the  pike;  thence  south  81  i° 
east,  114  perches  to  a  post ;  thence  south  2°  west,  62  perches  to  a  stone ; 
thence  south  88|°  west,  113f  perches  to  an  iron  stake  in  the  road; 
thence  north  2°  east,  80  perches  to  the  place  of  beginning,  containing 
50  acres  and  56  perches. 

6.  Beginning  at  a  corner  stone  in  the  middle  of  the  township  road ; 
thence  along  said  road  north  48°  west,  13  rods  and  9  links,  and  north 
20°  west,  32  rods,  and  north  6°  east,  54  rods  to  a  corner  in  said  road ; 
thence  north  71 1°  east,  138  rods;  thence  south  22°  east,  16  rods  to 
corner;  thence  south  45°  west,  171  rods  to  the  place  of  beginning,  con- 
taining 50  acres  and  38  rods  of  land,  be  the  same  more  or  less. 


Ill,  §  44] 


INDIRECT  MEASUREMENT 


61 


7.  Beginning  at  the  southwest  corner  of  fraction  24 ;  thence  N.  1  J°  E., 
33.23  ch.  to  a  stake  in  the  west  Hne  of  said  fraction;  thence  S.  89°  E., 
8.154  ch.  to  a  stone;  thence  N.  1^°  E.,  7.4  ch.  to  an  iron  pin  in  the 
north  boundary  line  of  said  fraction;  thence  S.  89°  E.,  15.436  ch.  to  a 
stake ;  thence  S.  32^°  W.,  13.50  ch. ;  thence  S.  35p  E.,  5.54  ch. ;  thence 
S.  48°  E.,  8.27  ch.  to  a  stake ;  thence  S.  7^  W.,  19.20  ch.  to  a  stake  in 
the  south  boundary  Une  of  the  lot;  thence  N.  89°  W.,  24.10  ch.  to  the 
place  of  beginning,  containing  85.87  acres,  be  the  same  more  or  less. 

44.   Relations  between  the  Sine,  Cosine,  Tangent.     In 

any  right  triangle  whose  sides  are  a,  b,  c,  Fig.  66, 

or,  dividing  by  c^, 

\2 


(!M 


But,  by  §  37,  this  is 

(1)  sia-A+cos^A  =  l. 

Since,   by   definition,   sinA=-,    cosA=-,   tanA  =  r,   it 

c  c  b 

follows,  by  dividing,  that 

sinil 


(2) 


C0Si4 


tani4. 


EXERCISES 

1.  Construct  a  right  triangle  whose  sides  are  3,  4,  5.     Find  the 
value  of  the  sine,  cosine,  and  tangent  of  each  acute  angle. 

Illustrate  formulas  (1)  and  (2) ,  §  44,  by  use  of  this  triangle. 

2.  Proceed  as  in  Ex.  1  for  a  right  triangle  whose  sides  are  8,  15,  17. 
For  one  whose  sides  are  7,  24,  25. 

3.  Proceed  as  in  Ex.  1  for  a  right  triangle  that  has  its  hypotenuse 
c  =  10,  and  an  acute  angle  =  53°  8'. 

4.  State  sin  A  in  terms  of  cos  A.     Also  cos  A  in  terms  of  sin  A. 

5.  Given  sin  A  =  .6.     Find  cos  A  and  tan  A. 

[Hint.     Construct  a  right  triangle  with  one  leg  6  and  the  hypotenuse 
10.    Find  the  other  leg.] 


62 


MATHEMATICS 


[HI,  §  44 


6.  Given  tan  A  =  1.     Find  sin  A  and  cos  A. 

7.  Given  tan  A  =  1.732     Use  a  table  to  find  sin  A,  cos  A. 

Show  that  the  following  relations  are  true  by  changing  one  or  both 
sides  of  the  relation,  making  use  of  (1),  (2),  §  44. 

8.  (sin  A+cos  A)2  =  l+2  sin  A  cos  A. 

[Hint.     Square  the  parenthesis  and  reduce  by  (1),  §  44.] 


9.    f-^+tanA'\  f^-r-tan a)  =1. 
\cos  A  /   \cos  A  J 


10.   cos  A 


tan  A  =  l. 


sin  A 

11.  sin2A(l+tan^A)=tan2A. 

12.  sin'*  A  —  cos^  A  =  sin^  A  —  cos^  A . 

13.  (sinA+cos  A)2+(sin  A-cos  A)2  =  2. 

^ .     1  —  sin  A  _    cos  A 
cos  A        1+sinA 

[Hint.     Clear  the  equation  of  fractions.] 

45.   Sine,   Cosine,   Tangent   of   Complementary  Angles. 
In  any  right  triangle,  Fig.  67,  A +  5  =  90°. 
By  definition. 


sin  A  =  -  =  cos  B, 
c 

A^^ 

^>-^ 

sinB=-=cos  A, 
c 

tanA-^      ,  ..  „• 

b 
Fig.  67 

h     tan  B 

Thus,  the  sine  of  an  acute  angle  equals  the  cosine  of  its 
complement,  etc. 

EXERCISES 

1.  Find  sin  50°,  cos  50°,  tan  50°  in  terms  of  functions  of  40°. 

2.  Express  each  of  the  following  as  a  function  of  the  complementary 
angle. 

(a)  sin  35°.  (c)   cos  47°.  (e)  tan  60°. 

(6)  tan  20°.  (d)  cos  75°.  (/)  sin  57°  24'. 


CHAPTER  IV 
REVIEW  OF  ALGEBRA 

46.  Use  of  Letters.  Historically,  the  first  use  of  algebra 
was  to  abbreviate  explanations  of  arithmetic  problems.  Its 
modern  use,  however,  is  much  broader.  It  includes  the 
free  use  of  symbols  to  state  rules  and  to  discuss  problems 
and  numbers  which  are  stated  in  words. 

For  example,  the  area  of  a  triangle  is  the  base  multiplied 
by  one  half  the  altitude.     This  may  be  stated  briefly  as 

A  =  i  ab, 

where  A  =  area,  a  =  altitude,  b  =  base. 

47.  Equations.  An  equation  is  a  statement  that  one 
expression  is  equal  to  another.  This  statement  is  written 
with  the  sign  of  equality,  = ,  between  the  expressions. 

Thus  the  equation  3  a;+4=  15  is  read,  3  x  plus  4  equals,  or 
is  equal  to,  15. 

If  a  value  of  an  unknown  quantity  is  found  which  makes 
the  two  sides  of  a  given  equation  equal,  it  is  said  to  satisfy 
the  equation.  This  value  is  called  a  solution  or  a  root  of 
the  equation. 

For  example,  if  a:+4  =  5,  the  value  x=l  satisfies  this 
equation,  since,  when  x=l,  both  sides  of  the  equation  are 
equal. 

To  determine  finally  whether  a  given  value  is  a  solution,  the 
value  should  be  substituted  in  the  equxition  to  see  if  it  does 
satisfy  it. 

63 


64  MATHEMATICS  [IV,  §  47 

Example  1.     Solve  the  equation 

3x+4  =  10/ 
Subtracting  4  from  each  side  of  this  equation  gives 
3x  =  10-4, 
or  3x  =  6. 

Dividing  each  side  by  3  gives 

x  =  2. 
This  is  the  solution  of  the  given  equation.     Substituting  2  for  x  in 
the  given  equation,  we  have 

6+4  =  10. 
This  means  that  the  equation  is  satisfied,  and  hence  2  is  a  solution. 
Example  2.    Solve  the  equation 

5x_       x_29 
6  4     12* 

Multiplying  both  sides  of  this  equation  by  12,  the  least  common 
multiple  of  the  denominators,  gives 

10  x -36  =  3  a: -29. 
Subtracting  3  x  from  each  side  and  then  adding  36  to  each  side,  we  find 
10x-3x  =  36-29, 
or  7  a:  =  7, 

whence  x  =  \. 

Substituting  a:  =  1  in  the  given  equation  gives 

oi  -V  =  -f|=-V. 

Therefore  x  =  1  is  a  root  of  the  equation. 

48.  Operations  on  Equations.  Without  disturbing  the 
equality  we  may 

(1)  Add  or  subtract  the  same  number*  on  each  side  of  the 
equation.  This  is  usually  performed  by  actually  moving  it 
to  the  other  side  of  the  equation  and  changing  its  sign. 

(2)  Multiply  or  divide  both  sides  of  the  equation  by  the  same 
number  *  except  that  division  by  zero  is  excluded. 

*  If  any  expression  other  than  a  simple  number  is  used,  it  must  at  least 
represent  a  number. 


IV,  §  48]  REVIEW  OF  ALGEBRA  65 

EXERCISES 

Solve  each  of  the  following  equations  to  find  the  value  of  the  unknown 
letter  or  term.     Check  each  solution. 

1.  2xH-3  =  7.        4.    -2a;+3=-7.      7.   12sinA+l=7. 

2.  2a;-3  =  7.        5.   7x-15  =  3x-7.  8.   2  tan  A +3  =  1+4  tan  A. 

3.  2a;+3=-7.    6.   6-5a  =  3-a.        9.   ^^x-\  =  2x-^. 

10.  The  number  of  cows  and  sheep  in  a  certain  farmyard  is  45,  and 
there  are  four  times  as  many  sheep  as  cows.     Find  the  number  of  each. 

11.  One  angle  is  twice  the  size  of  another  angle.  If  their  sum  is 
the  complement  of  three  times  the  larger  one,  find  each  of  the  two 
angles.  Solve  the  similar  problem  obtained  by  substituting  the  word 
supplement  for  the  word  complement. 

12.  A  does  f  of  a  piece  of  work  in  10^  days.  With  the  help  of  B 
they  then  finish  the  work  in  4  days.  How  long  would  it  take  B  alone 
to  do  the  work? 

13.  Find  the  angle  between  the  hands  of  a  clock  at  quarter  past 
four  o'clock.     At  what  time  will  they  be  exactly  together? 

14.  The  driver  of  a  milk  wagon  is  charged  $28.56  for  the  entire 
load.  He  carries  certified  milk  at  22^  per  quart  and  5  times  as  much 
milk  in  cans  at  16 ff  per  quart.     How  much  of  each  does  he  carry? 

15.  Find  four  consecutive  integers  whose  sum  is  94. 

16.  What  are  the  dimensions  of  a  right  triangle  whose  area  is  1.2 
acres  if  the  height  is  3.7  times  the  base? 

17.  In  a  triangle  one  exterior  angle  is  108°.  If  the  ratio  of  the 
other  two  angles  is  2.6,  find  the  three  angles  of  the  triangle. 

18.  A  miller  has  wheat  worth  $2.20  per  bushel  and  another  lot  worth 
$3.00  per  bushel.  He  wishes  to  mix  these  to  make  20  bushels  of  wheat 
which  shall  be  worth  $2.40  per  bushel.  How  much  of  each  shall  he 
take? 

19.  A  farmer  has  a  cow  whose  milk  contains  3^%  of  butter  fat 
(called  a  3|%  milk)  and  another  one  which  gives  a  5%  milk.  How 
shall  he  mix  them  to  obtain  50  lb.  of  a  4%  milk? 

20.  A  man  can  seed  a  wheat  field  in  8  days.  How  long  would  it 
take  him  to  do  it  with  the  aid  of  a  helper  who  can  do  only  |  as  much 
as  he  can? 


66  MATHEMATICS  [IV,  §  49 

49.  Addition.  In  business  and  in  ordinary  life  it  is  often 
necessary  to  add  positive  and  negative  numbers  or  expres- 
sions. To  add  8  x  and  —  5  x  is  merely  to  subtract  5  x  from 
8  X.    Longer  expressions  are  added  by  grouping  similar*  terms. 

Example.     Add  &x  —  %y-\-bz 

and  3a:+4  2/  — 6z 

the  sum  is  9x— 4?/—    z 

EXERCISES 
Find  the  sum  of  the  following : 

1.  2a+36-4c,  5c-4a-6,  and  6+5o-2c. 

2.  4  x-5  y-2  z,  —2  a;+3  y  —  7  z,  and  —9  x—y-\-^  z. 

3.  3A-5+2C,  5^-2B+C,  and -^+3B-4C. 

4.  5.6  x-2.7  y,  3.5  ar+4.2  y,  and  -2.2  a;-6.5  y. 

5.  5x2 -2.5  m,  -.5  a;2+4  m,  and  1.45  x2+3.2  w. 

50.  Subtraction.  To  subtract  one  quantity  from  another, 
we  proceed  as  in  addition  after  changing  the  sign  of  the  quantity 
to  he  subtracted. 

Example.     From  4a:+    y  —  l0z-\-2 

take  -x-6?/  +  ll2-3 


the  difference  is  5  x-\-7  y—21  z+5 

EXERCISES 

1.  From  6  a -7  6 -10  c  take  12  a+5  6 -6  c. 

2.  From  15X-6-2?/ take  —  5  2/4-9- a;. 

3.  From2-7A2+A  take3A-5A2+8. 

4.  From  1.5  J5-^2+2.8  take  .9^-3^2 _i, 

5.  From  20  r- 12.5  s-|-4.2  t  take  6  ^-2.4  r+9  s. 

6.  a2+24a-10-(16-5a+a2)  =  ? 

7.  x^-5x+7-{-2x^-5x+2)  =  ? 

8 .  Simplify  16—2  Vx + 5  Vx  —  24  —  7  Vx  by  combining  similar  terms. 


*  Similar  terms  are  those  that  have  a  common  factor.     For  example 
2x,  3x,  ax,  mnx  are  all  similar  with  respect  to  x. 


IV,  §  52]  REVIEW  OF  ALGEBRA  67 

51.  Removal  of  Signs  of  Grouping.  When  a  negative  sign 
precedes  a  parenthesis  or  other  grouping  -sign,  change  all 
positive  or  negative  signs  within,  including  the  sign  of  the 
first  term,  when  the  sign  of  grouping  is  removed.  But  when 
a  parenthesis  is  preceded  by  a  positive  sign,  no  change  is  made 
in  the  terms  when  the  grouping  sign  is  dropped. 

Example.  3x  —  (5x—2y)=3x—5x-\-2y. 

-(P+Q-i2)=  -P-Q+R. 

EXERCISES 

Remove  the  grouping  signs  and  add  similar  terms  in  each  of  the 
following  expressions. 

1.  8x-(7-2x). 

2.  ilQx-7  y)-ix-4:y). 

3.  (2p-5g)-h(4p+g)+6Q. 

4.  A-(3B-2.5)-(B-1.5A). 

5.  2a-(6-(2a-6)). 

6.  Solve  6  ?/ -  (4 -2  t/)  =20  for  y. 

7.  Solve  12.5x-(10-2x)  =  19-(x-2)  for  x. 

8.  Solve  7  x-[x-{5  a:-7)]+3  =  24  a;-9(H-x)  for  x. 

52.  Multiplication.  The  product  of  two  numbers  having 
like  signs  is  positive,  of  two  having  unlike  signs,  negative. 

For  example,  three  times  5  may  be  written 
(3)(5)  =  5+5+5=15. 
In  the  same  way  (3)(-5)  =  (-5)  +  (-5)  +  (-5)=  -15. 
The  multiplier  +3  is  here  a  positive  integer  and  the  multi- 
plicand is  added  three  times.     If  the  multiplier  is  negative, 
the  multiplicand  should  be  subtracted.     Thus 
(-3)(5)=-(+5)-(+5)-(+5)=-15, 
and  (-3)(-5)=-(-5)-(-5)-(-5)=-(-15)  =  +15. 

Thus,  generally, 

{a)(b)  =  ab  and      (-a)(-6)  =  a&, 

(a)(-6)=-a6     and         i-a)(b)=-ab. 


68  MATHEMATICS  [IV,  §  53 

53.  Exponents.  When  some  of  the  factors  of  a  number 
are  alike,  it  is  often  convenient  to  indicate  the  fact  as  follows  : 

2X4X4X4  =  2X43=128, 
3  a  ay  y  y  =  S  a^y^. 

The  small  figure  written  above  and  to  the  right  of  an  ex- 
pression denotes  the  number  of  equal  factors  and  is  called 
an  exponent. 

EXERCISES 
Multiply 

1.  4  0:2/  by  5  xy.  6.    arri^cd^  by  —a?md^. 

2.  2  a6c  by  fee.  7.    -4  aSxz/ by  -3  axV- 

3.  x^  by  x^.  8.   X*  by  xP. 

4.  3  fe^c  by  4  ahc^.  9.   a^  by  av. 
6.  7  r7  by  2  s^  by  3  rV.  10.  32»  by  Z^. 

Carry  out  each  of  the  following  indicated  multipUcations. 

11.  (a)  2(3+5).     (fe)  2{x^-y).     (c)  4a(a+6+ac). 

12.  a:(x2+l).  14.   a2(a2_a+i). 

13.  x3(2x4-3a:3+4x).  15.   x"(xi-«+x3+"). 

16.  Multiply  a:2+x2/+?/2  by  x-?/. 

17.  What  are  the  factors  of  a^  —  h^  as  shown  by  Ex.  16? 

18.  Multiply  a2-afe+fe2  by  a+fe. 

19.  Multiply  x'^+xy+y'^  by  x^— xi/+i/2. 

20.  Multiply  3a6-2a2-2aby  2afe-2a-3fe-6. 

54.  Standard  Type-forms.  Multiplications  may  fre- 
quently be  performed  mentally.  Some  standard  type-forms 
for  this  purpose  are  the  following : 

(a+fe)(a-6)  =  a2-62. 

(a+6)(a+6)  =  a2+2a6+62. 
(a-6)(a-b)  =  a2-2a6+fe2 

(a+6)3  =  a3+3  0^6+3  ab''+h\ 
(a-6)3  =  a3-3  a^fe+S  a62-6^ 
(a-f  6+c)2  =  a2+52_pc2+2  a?>+2  ac+2  6c. 
(a;+a)(a;+6)  =  x2+(a+6)a;-|-a6. 


IV,  §  56]  REVIEW  OF  ALGEBRA  69 

EXERCISES 

1.  State  in  words  the  type-forms  given  above. 

Carry  out  orally  each  of  the  following  indicated  multiplications. 

2.  {2x-3y)(2x-\-Zy).  11.    {a+b+2y. 

3.  (x-3)(x+3).  12.    {a-h-c)\ 

4.  ix^y-Sb){x^y-\-Sb).  13.    (2x-a+3)2. 

5.  (a+3  6)(a-3  6).  14.    (x+7)(x+2). 

6.  {x  +  [a+b])ix-[a+b]).  15.   (a-3)(a+5). 

7.  ix-2y)\  16.    (a:+2)(x-3). 

8.  {Sx+2yy.  17.    (3a6-x)(3a6+7x). 

9.  (x+2  2/)3.  18.    (p'q- 10)  (p'q+S). 
10.    (l-a)3.  19.    (x2-3)(x2-4). 

55.  Division.  The  rules  for  signs  hold  in  division  as  in 
multiplication,  as  shown  in  §  52. 

EXERCISES 

1.  Divide  y^  +  15  i/+36  by  2/ +4. 

2.  Divide  x^+x^-f?  x^-G  x+8  by  x2+2  x-f  8. 

3.  Divide  A3- 1  by  A-1. 

4.  Divide  a^-\-b^—c^-\-3  abc  by  a-\-b—c. 

6.    Divide  1+x  by  1  — x.     How  many  terms  in  the  quotient? 
6.    Divide  1  by  1  +a:  to  four  terms. 

56.  Factoring.  The  word  factor  means  ''  maker.*^  Fac- 
tors are  thus  merely  numbers  or  expressions  which  make 
other  numbers  or  expressions  when  multiplied  together. 
The  factors  of  6,  for  example,  are  2  and  3,  since  2  times  3 
is  equal  to  6. 

Factoring  is  the  process  of  finding  the  factors  of  an  ex- 
pression. Some  expressions  occur  quite  often  in  the  ap- 
plications of  algebra.  The  different  cases  may  be  grouped 
under  so-called  type-forms,  such  as  taking  out  a  common 
factor,  grouping  similar  terms,  the  difference  of  two  squares,  etc. 


70  MATHEMATICS  [IV,  §  56 

The  following  type-forms  will  aid  in  finding  the  factors  of 

many  expressions. 

ah-\-ac  =  a(b-\-c). 

a2-62=(a+6)(a-6). 
a'+2ah+¥={a+by. 
a'-2ah-\-¥={a-h)\ 

a^-h^=  (a-h){a'-j-ah-\-¥). 

a3+63=  (a-\-h){a^-ah+¥). 
x^-{-(a-\-h)x-\-ah=  (x-\-a){x-\-h). 

Examples.  A^-9  x^  =  (A+3x){A-3x). 

9  x2  -  30  xi/ +25  2/2  =  (3  a:  -  5  i/)2. 

EXERCISES 
Factor  each  of  the  following  expressions  by  inspection. 

1.  a%^-xK  7.   a:2+7a:  +  12. 

2.  (x+yy-zK  8.   a^+^ab-2l¥, 

3.  a2-(6+c)2.  9.   x3  +  l. 

4.  a^-(h-cy.  10.   x^-y\ 

5.  9x2+6x+l.  11.   8x3+27. 

6.  x2+4x2/+4?/2.  12.   a3-27  6«. 
Factor  each  of  the  following  expressions  by  grouping. 

13.  ax—ay-j-hx  —  by  =  a{x  —  y)+b{x—y)  =  {a+b){x—y). 

14.  2/3+3  1/2+3  1/  +  1.  15.   a^-ab^-a*b+bK 

16.  In  x2+6  x+k,  what  is  the  value  of  k  to  make  the  expression  a 
perfect  square? 

17.  How  much  must  be  added  to  x2 +8  x  to  make  it  a  perfect  square  ? 
How  much  to  x2 +5  X  to  make  it  a  perfect  square? 

Resolve  each  of  the  following  expressions  into  factors. 

18.  a^+Qa-\-S.  21.    1 -12  a6+35  a262. 

19.  x3-4x2+3x-12.  22.   48r2-10r-25. 

20.  3a4-8a2-35.  23.    15  p^-jyr-2r\ 

24.  4x2-13x+10. 


IV,  §  57]  REVIEW  OF  ALGEBRA  71 

57.  The  Binomial  Theorem.  From  the  following  equal- 
ities we  may  observe  the  law  for  raising  a  binomial  to  positive, 
integral  powers  by  inspection. 

(a+6)2  =  a2-|-2a6+62. 

(a+6)3  =  a3+3  a%-\-S  ab'-{-b\ 

(a+6)4  =  a4+4  a%+Q  a^¥-\-4:  a¥-^¥. 

{a-\-by  =  a'-\-5  a^fe+lO  a%^-\-10  a%^-{-5  a¥+¥. 

We  may  now  make  the  following  observations. 

(1)  The  number  of  terms  on  the  right  of  the  equality  sign 
is  always  one  more  than  the  exponent  on  the  left. 

(2)  The  exponent  of  the  first  letter  in  the  first  term  of  the 
expansion  is  the  same  as  the  power  on  the  left.  This  ex- 
ponent decreases  by  1  in  each  succeeding  term.  The  ex- 
ponent of  the  second  letter,  b,  in  the  second  term  is  1,  and 
increases  by  1  in  each  succeeding  term. 

(3)  The  first  coefficient  is  1 ;  the  second  coefficient  is 
exactly  the  power  to  which  we  are  raising  the  binomial; 
and  in  general,  the  coefficient  of  any  term  may  be  found  from 
the  preceding  coefficient  by  multiplying  it  by  the  exponent  of 
a  in  that  term  and  dividing  this  by  one  more  than  the  exponent 
of  b  in  that  term. 

For  example, 

(a+6)io  =  oio+10a96  +  ^^^  -0862+  ... 
1  *  ^ 

=  aio  +  10  0*6+45  a»b'-\-^^a''b^+  etc. 

o 

(a +6)"  =  a^ +na»-%  +  ^ '  ^.^  ~  ^^  a^-%^+  etc. 

These  facts  make  up  the  binomial  theorem.  It  is  true 
for  all  values  of  a  and  b  and  for  n  equal  to  any  positive 
integer.* 

*  The  binomial  theorem  may  be  extended  so  that,  under  certain  restric- 
tions, it  will  hold  for  other  values  of  n,  e.g.  fractional  and  negative  values 
of  n. 


72 


MATHEMATICS 


[IV,  §  57 


EXERCISES 

Expand  each  of  the  following  binomials  as  indicated. 

1.  {a+by.  4.    (a-2)3.  7.    (x-V^)'- 

2.  (a+6)".  5.    {a-2y.  8.   ^|  +  |)'- 

3.  (a+fe)i6.  6.    (x+Va)2. 

10.  Given  two  line-segments  a  and  b.  Con- 
struct a  square  on  the  segment  a +6.  Show  by 
means  of  this  square  that  {a+by  =  a^-\-b^+2  ab. 


Fig.  68 


58.  Fractions.  A  fraction  is  an  indi- 
cated division  of  one  number  or  expression 
called  the  numerator  by  another  number 
or  expression  called  the  denominator. 

As  in  elementary  arithmetic,  the  form  of  a  fraction  may  be 
changed  without  thereby  changing  the  value  of  the  fraction. 

The  value  of  a  fraction  is  unchanged  if  both  numerator  and 
denominator  are  multiplied  by  the  same  expression,  or  divided 
by  the  same  expression,  excepting  zero. 

Fractions  may  be  reduced  to  lowest  terms  by  dividing 
both  numerator  and  denominator  by  their  common  factors. 


EXERCISES 

Reduce  to  lower  terms  each  of  the  following  expressions. 

1    _6  2    —  3     ^!^.2-^.  4 

*  32'  '  725'  '     2^3  ab^  ' 


5. 


13. 


35  ax^y^ 
28  a^x^y 


a%-ab\ 
a^b^ 

15sin^a;tan3; 
20  sin  x  tan^  x 


x—y 

6s2-s-2. 
9s2-4 


10. 


6. 


-mp-Q)(r-sy(t-iy 

2S{p-q){s-r){l-t) 


^     ax+ay-x-y  ^^ 


t. 


X^-y^ 
X^-S 


x2-fx-6 


12. 


x^-4xy-{-3y^ 

sin^  X  —  cos^  X 
sin  x-f-cos  X 


14. 


12a^-aa;-6a^ 
8a2-2ax-3a2' 


IV,  §  60]  REVIEW  OF  ALGEBRA  73 

59.  Addition  and  Subtraction  of  Fractions.  In  order  to 
odd  or  subtract  fractions  they  must  he  reduced,  as  in  elementary 
arithmetic,  to  the  same  denominator.  This  denominator  should 
be  the  least  one  common  to  all  of  the  denominators, 

rru„„  2  ,5_14,  15     29 

^^^^  3+7-21  +  21  =21- 

Similarly,  a^x^^^^^g^, 

h     y      by      by  by 

EXERCISES 

Perform  the  operations  as  indicated  in  each  of  the  following  expres- 
sions. 


1. 

2^3^4 

9. 

X                X 

x-l      x+l 

2. 

3^15      V35     9; 

10. 

1+2^    3    . 
a     a^     a  +  1 

3. 

x-2  ,  2x+3 
3      '       2 

11. 

(a -6)2 

4. 

1+1.               6.    1- 
ah                      a 

1 

h 

12. 

3  1    5a     1        5 
2      a-2      (a-2)2 

6. 

*  +  1  • 

a+6      a—h 

13. 

5           7,1. 
x^-2     x-\      x^^x-1 

7. 

1            1 

a-\-h     a—h 

14. 

x-2     x-\  ,           3 
x-l      x-2     x2-3a:H-2 

8. 

X      ^     X     ^ 

x-l     x-\-\ 

15. 

0     5(a+6)  1  (a +6)2 
a-h        (a-6)2 

60.  Multiplication  and  Division  of  Fractions.  As  in 
arithmetic,  to  multiply  one  fraction  by  another  multiply 
their  numerators  together  for  the  numerator  of  the  product  and 
their  denominators  together  for  the  denominator  of  the  product. 
Any  factors  common  to  numerator  and  denominator  may  be 
cancelled  before  multiplying. 

To  divide  one  fraction  by  another,  invert  the  divisor  and 
then  multiply. 


74  MATHEMATICS  [IV,  §  60 

When  the  division  of  one  fraction  by  another  is  indicated  in 
the  form  of  a  fraction,  this  fraction  is  called  a  complex  fraction. 

a  2 

^,  b     a  ,  c      ad     ad  ^  3      2^5      5 

^^^'  -c  =  b—d  =  -b''-c  =  Vc'  6=3><6=9' 

d  5 

EXERCISES 

Perform  each  of  the  following  indicated  operations. 

■'■  (i--:)x-L- 

\        9^/      x—y 


3^4^7^9 


^     35     ^-39a  ^    V^x^-^^r^^^i 


15  g^.  -117  .,    18  6 

27  62 


3.    "^^x ^ .  9     15^'^25_5r. 

2         x2-2x-3  '      32r     '     36 

x'^-^y^  10      4x2    .   -12  ax 


x-^ .     ^  >  10. 


x  —  2y       x^—y^  —ay"^  9y 

5      r2^6t;+9  ^  t;-4^  ^^       2  <2    .     < 

16-8y+y2      9-y2'  '    a'+i^  *.  a_^^' 


Simplify  the  following  expressions. 


J  ,  a  -.  _  2jc  _j_  £2 


3  6 

4 

14.    Z.  16.    £±i-  18. 

^  111 


5 


1+- 


1- 

x2 
a2 

x+y 
x 

y 

x—y 

X 

x+y 
y 

61.  Simple  Linear  Equations  of  One  Unknown  Quantity. 
A  linear  equation  is  one  in  which  there  is  no  term  of  higher 
degree  than  the  first. 

We  shall  now  study  some  equations  of  the  first  degree 
which  are  more  complicated  than  those  given  earlier  in  this 


IV,  §  61]  REVIEW  OF  ALGEBRA  75 

chapter  but  which  may  be  solved  by  the  same  methods 
used  there. 

Equations  may  be  cleared  of  fractions  by  multiplying 
both  sides  of  the  equation  by  the  least  common  multiple 
of  the  denominators. 

Example.     Solve  the  following  equation  for  x : 

6a;-3  ^3x-2 
2x4-1  X 

The  equation  is  cleared  of  fractions  by  multiplying  both  sides  by 
the  least  common  denominator,  (2  x-\-l){x).     This  gives 

x(6a:-3)  =  (3x-2)(2j:-fl), 
or 

6x2-3x  =  6x2-x-2. 

Collecting  the  like  terms,  we  find 

-2x=-2 
and  thus 

x  =  l. 

The  solution  may  be  checked  by  substituting  1  for  x  in  the  original 
equation. 

EXERCISES 
Solve  each  of  the  following  equations. 

1.4(2.-1)-6(|-|)=.2 

2.  4(x-|)2+6(x-f)  =  (2x-5)2+302. 

3.  One  side  of  a  right  triangle  is  5  inches  and  the  other  side  is 
2  inches  shorter  than  the  hypotenuse.     How  long  is  the  hypotenuse? 

4.  When  the  radius  of  a  circular  plot  of  ground  is  increased  2  feet, 
the  area  is  increased  88  sq.  ft.     Find  the  size  of  the  original  plot,  using 

7r  =  22/7. 

5.  At  what  time  between  3  and  4  o'clock  are  the  hour  and  minute 
hands  of  a  clock  together? 

6.  Calculate  when  you  were,  or  will  he,  half  as  old  as  your  father. 


76  MATHEMATICS  [IV,  §  61 

Solve  each  of  the  following  equations,  the  last  letters  of  the  alphabet 
X,  y,  z,  w,  etc.  being  the  unknowns. 

7.    ^+3  =  4.  3+?^^ 

4  3  4 


^^+2      ^  11.   ax-j-bx  =  c-\-d. 

9.    (a+6)(x+2)=c.  12.   y  =  ^%hll. 

0 

13.  (a+2)(fc-0)-(a-z)(c+z)=2. 

14.  Solve  the  equation 

l+i 
for  i.     Also  solve  it  for  I. 

15.  Solve  the  following  equation  for  F : 

C  =  i(F-32). 

This  equation  represents  the  relation  between  the  readings  on 
Centigrade  and  Fahrenheit  thermometers.  What  is  the  Centigrade 
reading  for  68°  Fahrenheit?     Change  25°  C.  to  Fahrenheit  reading. 

16.  A  farmer  made  100  lb.  of  fertilizer,  using  95  parts  acid  phos- 
phate, 100  parts  cottonseed  meal,  and  5  parts  muriate  of  potash.  He 
then  decided  to  add  enough  acid  phosphate  to  each  100  lb.  of  the 
original  mixture  to  make  the  acid  phosphate  content  50%  of  the  total. 
How  much  should  he  add  ?  Ans.     5  1b. 

[Hint.     Let  x  =  the  number  of  pounds  added.     Then 
50%  of  (100  +  x)  =47^  lb. -j-x.]. 

17.  In  Ex.  16,  how  many  pounds  less  of  acid  phosphate  should  be 
used  to  every  15  pounds  of  muriate  of  potash,  in  order  to  reduce  the 
acid  phosphate  content  of  the  mixture  to  40%? 

18.  In  the  fertilizer  of  Ex.  16,  how  many  pounds  less  of  acid  phos- 
phate shall  be  used  to  every  60  pounds  of  cottonseed  meal  in  order  to 
reduce  the  acid  phosphate  content  of  the  mixture  to  42.5%? 

19.  A  fertilizer  used  in  cotton  fields  consists  of  62^%  acid  phos- 
phate, 30%  dried  blood,  and  7^%  muriate  of  potash.  It  is  desired  to 
raise  the  acid  phosphate  content  to  70%.  How  many  pounds  of  acid 
phosphate  must  be  added  to  one  ton  to  do  this? 


IV,  §  62]  REVIEW  OF  ALGEBRA  77 

62.    Quadratic   Equations.     A   quadratic   equation   is    one 

in  which  there  is  a  term  of  the  second  degree,  but  none  of 
higher  degree.     There  may  be  terms  of  lower  degree. 
For  example, 

is  a  quadratic  as  it  contains  terms  of  the  second  and  lower 
degree. 

Quadratics  may  be  solved 

(a)  by  factoring,  or 

(6)  by  completing  the  square. 

Example.     Solve  the  equation  3  x^+G  =  11  a:  by  factoring. 
Transpose  all  terms  to  the  left  side  of  the  equation.     Thus 

3x2-llx+6  =  0. 
Factoring  the  left-hand  side,  we  obtain 
(3x-2)(x-3)=0. 

But  this  means  that  each  factor  may  be  put  equal  to  zero,  since  a  product 
is  zero  if  any  of  its  factors  are  zero.     Thus  either 

3a:-2  =  0,  i.e.  x  =  f, 
or  else 

X— 3  =  0,  i.e.  x  =  S. 

Check :   Substituting  f  for  x  in  the  original  equation  gives 

3.f+6  =  ll.f,  or  V+6=¥- 

Substituting  3  for  x  in  the  same  equation  gives 

3-  32+6  =  33. 

EXERCISES 

Solve  each  of  the  following  quadratics  by  factoring. 

1.   5x2+6x  =  32.  2.   3x2-44x=-121. 

3.   4x2+5x  =  0.  4.   x2-2  6x  =  2ax-4a6. 

5.  3x2  =  12. 

Notice,  in  this  example,  the  double  sign  obtained  on  extracting  the 
square  root. 

6.  11x2-10  =  5x2+44.  7.    (2x+3)2  =  5x+39. 


78  MATHEMATICS  [IV,  §  63 

63.   Solution  of  the  Quadratic  by  Completing  the  Square. 

The  method  of  solution  in  this  case  may  be  observed  in  the 
following  examples. 

Example  1.  Solve  the  equation  ^2+27  =  12  x  for  x  by  completing 
the  square. 

Transpose  the  terms  containing  x  and  x^  to  the  left  side  of  the  equa- 
tion, those  not  containing  an  x  to  the  right  side  of  the  equation.  Thus, 
x^-\2x=-27. 

To  make  the  left  side  a  perfect  square  we  must  take  half  oj  12,  square  it, 
and  add  it  to  both  sides  of  the  equation.     This  gives 

a:2- 12  a: +36  =-27+36  =  9, 
or 

(x-6)2  =  9, 
whence 

Therefore 

a:  =  6=«=3,     i.e.     x  =  9     or    x  =  S. 

Example  2.     Solve  the  equation  7  x  =  3x^—6. 
Transposing  the  unknown  terms  to  one  side  and  the  constant  to  the 
other  side,  we  find 

3a:2-7x  =  6, 
or,  dividing  by  3, 

x^-^x  =  2. 

Half  the  coefficient  of  x  is  —  ^.  If  we  square  this  and  add  it  to  each 
side,  we  obtain 


or 


(x-|r=W- 

Thus  a;-|==b-V, 


whence 

x  =  i-\-^-,     or     x=i—^, 
i.e.  x  =  3     or    x=— |. 

EXERCISES 

Solve  each  of  the  following  equations  by  completing  the  square  and 
check  the  answers. 

1.  2x2+4  =  9x.  4.   8p2_2p-15  =  0. 

2.  3x2+5a:-12  =  0.  5.   5  tan^  A -8  tan  A  =21. 

3.  4  22-20  2+25  =  0.  6.    12  sin^  A +6  =  17  sin  A. 


IV,  §  64]  REVIEW  OF  ALGEBRA  79 

64.  Simultaneous  Equations.  Two  (or  more)  distinct 
equations  which  contain  the  same  unknown  quantities  are 
called  simultaneous  equations,  if  they  are  considered  at 
the  same  time. 

Solution  of  simultaneous  equations  is  possible,  in  general, 
only  when  the  number  of  given  equations  is  the  same  as  the 
number  of  unknown  quantities. 

Example  1.  The  cost  of  a  bushel  of  potatoes  and  a  bushel  of  apples 
was  $4.00,  but  the  apples  cost  20ff  more  than  the  potatoes.  Find  the 
price  of  each. 

Let  a; = the  price  of  a  bushel  of  apples, 

and 

Then 

and 

Adding  gives 


y=the  price  of  a  bushel  of  potatoes. 
a;+y  =  400, 
x-y  =  20. 


2x  =  420, 
x  =  $2.10  the  price  of  apples, 
whence  2/ =  $1.90  the  price  of  potatoes. 

Either  of  the  above  equations  taken  singly  would  not  be  sufficient 
to  determine  x  and  y. 

Example  2.     Solve  the  simultaneous  equations 
\2x-  5?/  =  60, 
\7x-\2y  =  M2. 

Multiply  the  first  equation  by  7,  the  second  one  by  2.    Thus 
14  a: -351/ =  420, 
14x-24i/  =  684. 

Now  subtract  and  we  have 

-111/ =-264,     or    2/ =  24. 
For  y  in  the  first  equation  put  24,  giving 
2  X- 120  =  60,     i.e.     re  =  90. 

Thus  the  set  of  values  x  =  90,  i/  =  24  satisfies  the  second  equation 
and  is  said  to  be  the  solution  of  the  set  of  equations. 


80 


MATHEMATICS 


[IV,  §  64 


EXERCISES 
Solve  each  of  the  following  sets  of  simultaneous  equations. 

3  A +2  5  =  32, 


1. 


2. 


|5o+6  =  21, 
[2a+3&  =  24. 


3. 


{Sx+7y= 
\l2x-5y 


=  100, 

88. 


20A-3B  =  1. 
3x 


19 

2x 


+5  2/ =  13, 
-33  =  ^^-^. 


5.  The  standard  daily  ration  for  an  adult  laboring  man  should  be 
about  16  oz.  starch,  4  oz.  fat,  and  4  oz.  albumen.  These  materials 
as  found  in  bread,  butter,  and  beef  are  as  follows. 


Food 

Starch 

Fat 

Albumen 

Bread 

Butter 

Beef 

54% 
0% 
0% 

1% 

83% 
15% 

9% 

1% 
15% 

Find  the  quantities  of  these  foods  required  to  make  a  daily  ration. 

6.  If  a  dairy  cow  requires  1  lb.  of  protein  to  6  lb.  of  carbohydrates 
in  her  food,  what  should  be  the  proportions  of  dry  peas  and  hay  for  her 
daily  ration  if  the  dry  peas  contain  10  lb.  of  protein  and  32  lb.  of  carbo- 
hydrates per  bushel  of  60  lb.,  and  the  hay  88  lb,  of  protein  and  880  lb. 
of  carbohydrates  per  ton  ? 

65.  Fractional  Exponents.  Thus  far  we  have  used  only 
exponents  that  are  positive  integers.  We  defined  a^  to 
mean  aa,  a"   =aaa  •••  to  n  factors,  etc. 

It  is  often  convenient  to  use  fractional  exponents,  such  as 
x^/^.  But  X  taken  as  a  factor  f  times  has  no  meaning.  Since 
it  is  important  that  the  symbols  used  should  always  follow 
the  same  rule,  we  shall,  in  the  case  of  exponents,  extend  the 
definition  for  combining  them  so  as  to  include  fractional  and 
negative  values.     That  is,  we  will  make  the  relation 

always  true,  whatever  p  and  q  may  be. 


IV,  §  67]  REVIEW  OF  ALGEBRA  81 

Thus  o}''^  shall  mean  a  number  such  that 

Therefore,  a}'"^  is  the  square  root  of  a  and  a^''^  =  Va. 
Likewise  a^^*  shall  mean  that 

fl3/4   .  ^3/4   .   fl3/4   .   a^/^  =  (^, 

i.e.  a^^*  IS  </ie  fourth  root  of  o?  and  a^/*=  'V^a^. 

Thus,  generally,  a^^*^  shall  mean  a  number  such  that 
flP/*/  .  aP/q  .  aplq  ...  (<o  q  factors)  =  oP, 

i.e.  aPi"^  is  the  qth  root  of  a"  and  aP"^=  Va^. 

Therefore,  in  a  fractional  exponent,  the  numerator  denotes 
a  power,  the  denominator  denotes  a  root. 

Examples.     ai/3  =  v/a;       a^/&  =  VS^^  =  {VaY  ]       x^/^  =  y/x^] 
82/3  =  ^==^64  =  4,     or    82/3  =  (8V3)2  =  (2)2  =  4. 

66.  Zero  Exponent.  To  find  the  meaning  of  aP.  From 
the  previous  paragraphs 

a^  •  a"  =  a^+"  =  a",       whence      a®  =  c"  -^  a"  =  i . 
Thus,  any  number  whose  exponent  is  zero  is  equal  to  1. 
Examples.    a:«  =  1,  10°  =  1,  (1)"  =  1,  etc. 

67.  Negative  Exponents.  To  find  the  meaning  of  a~" 
{n  positive).     From  the  previous  paragraphs, 

a""a"  =  aO=l,       whence,       a"  =  — • 

Therefore,  to  change  the  sign  of  the  exponent  of  a  number, 
take  the  reciprocal  of  the  number,  and  affix  the  exponent  with 
its  sign  changed. 

Examples.     x-^  =  —  ;      x-'/2  =  -^  =  -J-  ;       J-  =x^ : 
X'  a:i/2      Vx  2;-2 

— i— =a;3/2  =  V^;       a+x-^  =  a  +  -:       8-2/3  =  J_. 

X-3/2  '  '   X  82/3 


82  MATHEMATICS  [IV,  §  68 

68.  Definitions.     By  definition,  therefore, 

(1)  a"  =  aaa    •to  n  factors  if  n  is  a  positive  integer. 
Thus  a^  =  aa, 

(2)  a^^'=  </a,  i.e.  the  rth  root  of  a,  as  a^/^  =\^a. 
a"A=  Va"  =  the  rth  root  of  a", 

=  (ai/'')"=(^a)«. 
Thus 

^2/8  =-^  ^  (^)2 .  253/2  =  V25'  =  ( V25)3  =  125. 

(3)  aO=l. 

(4)  a-"  =  —  (n  positive  or  negative,  fractional  or  integral). 

a" 

69.  Laws  of  Exponents.     For  n  and  r  any  positive  or 
negative  number,  fractional  or  integral, 

(1)  a'"  •  a''=a'"+'', 

a' 

i.e.    add   exponents   when   multiplying,    subtract   them   when 
dividing  powers  of  quantities. 

(2)  (a")'-=a"', 

i.e.  multiply  exponents  in  raising  to  powers j  divide  the  ex- 
ponents in  extracting  roots. 

(3)  a"'b'^={ab)\ 

EXERCISES 

Write  each  of  the  following  expressions  without  zero  or  negative 
exponents. 

1.   a-3.      2.   a^x-K      3.    ^^t^      4.    ^^^^ 


2x-3  2a-i6o  100x-i-a:-2 

6.   Evaluate  each  of  the  following  expressions. 

163/4;     (-8)2/3;     (25-l)-3/2;     (^^  ^' ;     (^)''^';     (27-^)2/3. 


IV,  §  70]  REVIEW  OF  ALGEBRA  83 

7.  Write   each   of    the   following   expressions  with   radicals   and 

positive  exponents. 

0x1/3;    (ax)i/3;  322/S;  32-2/5;    21  x-3/2y2  ^    ^g  ^2^_3)2/3. 

3  x~hj~^ 

8.  Write  each  of  the  following  expressions  with  fractional  exponents 
instead  of  radicals,  and  vice  versa. 

vT^;   y/hi\   v^iV^;    {^^f^\"^]  2:2/32^2/3;   (5-1/2x2/363/2)6. 

9.  Simplify 

V25x^a%-^;    v^277W;   V3  a«+6  afc+3  6^;    v^(x+2/)Ha;+2/)~2/3. 
Perform  the  indicated  operations  in  each  of  the  following  expressions. 
10.    (x2/3 +2/2/3) (x2/3- 2/2/3  ).  h.    (a*^a^^l)(a-*-a-^+l). 

12.  (x-2/3+x-l/3-fl)(x-l/3-l). 

13.  (21  a-|-a2/3+ai/3  +  l)--(3ai/3  +  l). 

14.  (x-3+2x-2-3x-l)-^(x-2+3x-l). 

Add  a  third  term  to  complete  the  square  in  each  of  the  following 
expressions. 

15.   x+2xi/2  +  ?  16.   9+6ai/2  +  ?  17.   a-2+8a-i  +  ? 

70.  Reduction  of  Radicals.  Similar  Radicals.  The  rules 
for  operating  with  radicals  follow  from  the  preceding  para- 
graphs. Radicals  are  said  to  be  similar  if  they  differ  only 
in  their  coefficients. 

A.  To  introduce  a  quantity  under  the  radical  sign,  raise 
the  quantity  to  the  form  of  the  radical  and  then  multiply 
them  together. 

Examples.    5's/2  =  V25.  V2  =  V50. 
2^2=^.^=^l6. 

B.  To  reduce  a  radical  to  a  simpler  form,  factor  the  quantity 
under  the  radical  sign  into  factors  so  that  the  indicated  root 
may  be  performed  on  them. 


84  MATHEMATICS  [IV,  §  70 

Examples.     Vl2  =  V(4K3)  =  V4V3  =2V3. 
v/54=v/27:2  =  3v/2. 

EXERCISES 

Simplify  each  of  the  following  expressions  by  taking  out  a  factor 
from  under  the  radical  sign. 

1.    V288.  2.    V98O.  3.    V75O.  4.    V49(a+6)3. 

71.  Addition,  Subtraction,  Multiplication,  Division  of 
Radicals.     Similar  radicals  may  be  added  or  subtracted. 

Example.     2V2O  + V27- Vi  +  \/45  =  4V5+3V3-iV3+3V5 

=  7V5+|V3. 

Similar  radicals  may  also  be  multiplied  or  divided  by  per- 
forming these  operations  on  the  coefficients  and  on  the 
quantities  under  the  radical  signs  since 

a^h^={ahy  and  ^  =  f-Y. 
6«     \bj 

EXERCISES 

Simplify  each  of  the  following  expressions. 

1.    V6  +  V24-3V54+3V6.  2.   ^;  ^|;  ^|;  yjl^,. 

3.  VI8  +  V5O-V75  +  V48  +  V98-V32. 

4.  V26VI5VJ  Ans.     12 \^. 

5.  3V54V24-^Vl50.  7.    VI2 Vl5 -- V24 Ve. 

8  ^y 

72.  Small  Errors.  Numbers  obtained  from  actual  meas- 
urements are  subject  to  an  error  of  not  more  than  5  in  the 
next  place  after  the  last  one  written. 

For  example,  if  a  measured  length  is  reported  to  be  3.47', 
we  understand  the  true  length  to  be  somewhere  between 


IV,  §  74]  REVIEW  OF  ALGEBRA  85 

3.465'  and  3.475'.  If  the  last  figure  is  0,  it  means  that  the 
reading  is  rehable  in  that  place.  For  example,  6.20'  means 
that  the  true  value  is  somewhere  between  6.195'  and  6.205'. 
This  applies  to  those  measurements  which  are  stated 
without  any  indication  of  the  probable  error  involved. 

73.  Error  in  a  Sum.  Suppose  a  and  b  are  the  readings  of 
two  measurements  and  that  ei  and  62  are  their  errors,  re- 
spectively. The  true  values  are  therefore  a=tei  and  6  ±62. 
Their  sum  is  therefore 

(a=fcei)4-(6±e2)  =  (a+6)±ei±e2. 

The  sum  of  the  two  readings,  a +6,  may  thus  have  an  error 
of  not  more  than  ei+^2.    This  may  be  expressed  by  saying : 

The  error  of  a  sum  is  the  sum  of  the  errors. 

Example.  The  sides  of  a  field  measure  127' ±.2',  231'=*=  .4',  and 
159'  =*=  .3'.     Find  the  possible  error  in  the  sum  of  the  sides. 

ei  =  .2',  62  =  A',  63  =  .3'  whose  sum  =  .9'. 
127+231+159  =  517. 

The  true  value  of  the  sum  is  thus  between  517' +  .9'  and  517'  — .9'. 

74.  Error  in  a  Product.  Suppose  a  and  b  are  two  meas- 
urements whose  errors  are  61  and  62,  respectively.  The 
product  of  the  two  magnitudes  is  then  possibly 

(a  =1=  61)  (6  =«=  62)  =  a&  =•=  ae2  =*=  661  =«=  6162. 

The  error  of  the  product  is  thus  not  more  than  ae2-\-bei-^eie2. 
But  as  61  and  €2  are  both  small  as  compared  with  a  and  6, 
their  product  61^2  is  so  small  correspondingly  that  it  may  be 
neglected.     The  error  in  ab  is  thus  very  well  represented  by 

ae2-\-bei. 

Thus,  the  error  of  a  product  ab  is  a  times  6's  error  plus  b 
times  a' 5  error* 


86  MATHEMATICS  [IV,  §  74 

Example.  Find  the  extreme  error  in  the  computed  area  of  a  rectan- 
gular field  whose  sides  are  found  by  measurement  to  be  (127'  =*=  .2')  and 
(231' =±=.4'). 

The  extreme  error  made  in  taking  the  product  127  X  231  would  be 
(127)  (.4) +  (231)  (.2)  =97  sq.  ft. 

The  true  value  of  the  product  of  the  two  given  numbers  is  therefore 
somewhere  between  45,507+  97  and  45,507  —  97,  in  square  feet.  It  is 
remarkable  that  so  great  an  uncertainty  in  area  may  result  from  such 
small  errors  in  the  original  measurements. 

75.  Error  in  a  Quotient.  The  quotient  of  two  measure- 
ments, a  and  6,  whose  errors  are  ei  and  62,  respectively,  is 

b=i=e2     b       6(6^62) 

Since  62  is  small  as  compared  with  6,  we  may  replace  6=«=e2 
in  the  denominator  by  b.  The  error  in  the  quotient  is  then 
very  well  represented  by 

ae2-\-bei 

Thus,  the  error  in  a  quotient  -  is  a  times  b^s  error  plus  b 

b 

times  a's  error,  divided  by  the  square  of  b. 

Example.     The  error  in  the  quotient  of  625'  ±  .7  and  36'  =fc  .2'  is 

(36)  (.7) +  (625)  (.2)  _  ^^ 
362 
The  true  value  of  the  quotient  may  therefore  be  between 

W  +  -12andW--12 

EXERCISES 

1.  Add  (3.61  ±0.04)  and  (5.84  ±0.03)  and  find  the  possible  error  in 
the  sum. 

2.  What  is  the  extreme  error  in 

(55.6  ±  0.4)  +  (71 .8  ±  0.5)  -  (4.3  ±  0.3)  ? 

3.  Find  the  extreme  error  in  the  product  of  (351  ±0.2)  and  (127  ±  .7). 
Test  the  validity  of  the  answer  by  multiplying  together  the  extreme 

possibilities,  viz.  (351.2)  times  (127.7)  and  (350,8)  times  (126.3). 


IV,  §  75]  REVIEW  OF  ALGEBRA  87 

4.  Compute  the  error  in  (1672  ±0.4)  times  (896  ±0.5). 

5.  Compute  the  errors  in  the  following  quotients. 

(a)  136  ±0.3  ^52  ±0.8 
(6)  526  ±0.7  ^36  ±0.4 
(c)   7.321  ±0.018  ^1.234  ±0.009 

6.  If  Z  is  the  measured  length  of  the  side  of  a  square  field  and  if  e 
is  the  small  error  to  which  it  is  liable,  find  the  total  error  in  the  computed 
area  of  the  field.     Also  the  approximate  error. 

[Hint.    The  area  is  (i±e)2.] 

7.  What  is  the  area  and  its  possible  error  for  a  square  field  measur- 
ing 247'  if  the  measured  side  is  liable  to  an  error  of  0.5'? 

8.  The  diameter  of  a  circular  plate  is  6".  Find  the  increase  in 
its  area  due  to  heating  if  the  radius  has  thereby  been  lengthened  0.1". 
What  does  this  mean  geometrically? 

9.  What  is  the  error  in  the  volume  of  a  cube  whose  side  is  a  when 
the  side  is  liable  to  an  error  e  in  its  measurement? 

[Hint.  The  volume  is  (a-\-ey  =  a^+Sa^e+Sae^+e^.  Which  of 
these  terms  may  be  neglected?  Why?  What  then  is  the  error  in 
the  volume?] 

10.  The  side  of  a  cube  measuring  2.1"  on  each  side  is  increased 
0.02"  due  to  heating.     What  is  the  corresponding  increase  in  volume? 

11.  Draw  a  square  to  represent  (4.01)  (4.01).  Indicate  thereon  the 
meaning  of  0.01  X  0.01. 

Do  the  same  for  a  rectangle  whose  sides  are  4.01  and  7.06.  Show 
also  the  meaning  of  (4)  X(0.06).     Of  (7)  X(O.Ol).     Of  4  X  7. 

12.  What  possible  inaccuracy  may  exist  in  the  area  of  a  rectangle 
whose  measured  sides  are  22.34'  and  16.47'?  Find  the  area  of  the 
rectangle. 

[Hint.  The  error  in  22.34  may  be  as  great  as  0.005'.  Thus  the  sides 
may  be  written  (22.34' ±0.005')  and  (16.47' ±0.005').] 

13.  If  a  bin  containing  wheat  is  approximately  6'  X  15'  X  4',  show 
that  an  error  of  Hn.  in  measuring  each  of  the  dimensions  would  make  a 
difiference  in  value  of  more  than  $15.00,  if  wheat  is  $2.20  per  bushel. 


CHAPTER  V 
GRAPHIC  REPRESENTATION  OF  QUANTITIES 

76.  Positive  and  Negative  Numbers.  Directed  Line 
Segments.  Scales.  So  far  we  have  considered  a  line 
segment,  as  OX,  Fig.  69,  to  have  length  only.  Its  measure 
has  always  been  a  positive  number.     However,  it  is  often 

-8-7-0-5-4-3-2-1    012345678 
A'  O  A 

Fig.  69.  —  Positive  and  Negative  Directions  on  a  Marked  Line 

of  very  great  convenience  to  make  use  of  the  notion  of 
positive  and  negative  numbers.  This  is  desirable  when 
we  wish  to  distinguish  between  the  two  opposite  directions 
or  senses  which  exist  on  a  line.  Let  0  be  any  fixed  point  of 
reference  on  a  straight  line,  as  in  Fig.  69.  The  position  of 
any  other  point,  as  A  or  A' ,  is  known  as  soon  as  we  know 
its  distance  from  0  and  its  direction  from  0.  We  may  thus 
say  that  A  is  3  units  to  the  right  of  0  and  that  A'  is  3  units 
to  the  left  of  0.  If  we  agree  that  segments  measured  to  the 
right  of  0  shall  be  positive  and  those  to  the  left  of  0  shall 
be  negative,  we  may  say  that  A.  is  +3  units  from  0  and  that 
A'  h  —3  units  from  0. 

Such  a  marked  line,  showing  the  positive  and  negative 
numbers  representing  the  points  in  regular  order  to  the 
right  and  left  of  the  origin,  respectively,  is  known  as  a 
number-scale  or  an  algebraic  scale.  A  foot  rule,  or  a 
yardstick,  or  a  centimeter  scale  are  examples  of  scales. 
They  show  only  positive  numbers.  A  thermometer  is  a 
good  illustration  of  a  vertical  scale  that  shows  numbers 
above  and  below  zero,  for  example,  10°  and  — 10°. 

88 


V,  §  77]  GRAPHIC   REPRESENTATION  89 

The  directed  segment,  used  as  a  unit  on  scales,  is  not 
necessarily  a  straight  line.  For  example,  a  protractor  is  a 
semicircular  scale,  §  1,  while  many  greenhouse  temperature 
charts  and  weather  statistics  are  registered  on  paper  fastened 
on  revolving  cylinders. 

EXERCISES 

1.  A  bicylist  starts  from  a  certain  point  and  rides  +18  mi.  and 
then  —10  mi.  on  a  straight  road.  How  far  from  and  on  which  side  of 
the  starting  point  is  he? 

2.  How  far  and  in  what  direction  from  the  starting  point  is  an  auto 
which  goes  eastward  and  westward  according  to  the  following  numbers : 
+  16  mi.,  then  —6  mi.,  then  —20  mi.,  then  +28  mi.,  then  —15  mi.? 

3.  Give  the  meaning  of  the  following  latitudes,  using  the  equator 
as  the  origin:  +28°,  -18°,  +2°,  -23°.  (What  large  city  in  the 
western  hemisphere  is  in  latitude  —23°?     What  one  in  latitude  +23°?) 

How  are  places  located  on  the  earth's  surface?  Give  a  compre- 
hensive answer. 

4.  What  do  -4°,  0°,  +100°  mean  on  a  Fahrenheit  thermorr.eter? 
On  a  Centigrade  thermometer? 

5.  What  is  the  significance  of  points  to  the  left  of  the  origin  when 
the  line  represents  your  bank  account  ? 

6.  Compare  by  use  of  line  segments  the  amount  of  butterfat  in 
the  milk  produced  by  the  three  types  of  cows  mentioned  in  Ex.  9,  §  8. 

77.   Graphs  of  Data.     Axes  of  Reference.     We  are  now 

enabled  to  represent  graphically  various  kinds  of  informa- 
tion implying  directly  or  indirectly  positive  and  negative 
values.  Such  representations  are  used  largely  in  locating 
points  on  a  diagram,  or  in  locating  places  as  in  geography, 
or  in  numbering  houses  in  cities.  This  is  done  with  reference 
to  two  fixed  lines  or  axes,  usually  at  right  angles  to  each 
other.  In  cities  two  main  streets  are  used  as  axes  of  refer- 
ence, in  geography  the  equator  and  the  prime  meridian  are 
so  used,  while  in  more  local  geography  township  lines,  main 
roads,  etc.,  are  assumed  as  axes  of  reference. 


90 


MATHEMATICS 


[V,  §  78 


78.  Comparison  of  Data.  This  same  system  may  be  used 
to  compare  two  sets  of  quantities  or  data  where  a  value  in 
one  set  depends  upon  or  corresponds  to  a  value  in  the  other  set. 
Examples  of  frequent  occurrence  are  time  of  day  and  the 
temperature  in  degrees,  the  age  of  a  calf  and  its  weight,  the 
distance  traveled  by  an  auto  and  the  number  of  hours  since 
starting.  Information  of  this  kind  can  be  exhibited  on  a 
diagram  that  shows  the  pairs  of  corresponding  values.  To 
do  this,  squared  paper,  or  rectangular  coordinate  paper,  is 
usually  used.  This  is  merely  paper  ruled  into  small  squares 
or  into  small  rectangles.  Two  scales  or  axes  at  right  angles 
are  chosen  —  one  for  each  set  of  data  to  be  compared. 
Pairs  of  related  values  of  the  data  are  represented  by  points 
which  are  located  with  reference  to  these  two  axes.  By 
connecting  these  points  taken  in  order,  a  curve  is  obtained 
showing  the  different  values,  their  changes  from  one  value 
to  another,  etc. 


Example.  On  a  winter's  day  the  thermometer  was  read  at  6  a.m. 
and  every  hour  afterward  until  5  p.m.  The  readings  were  :  — 10°,  —8°, 
-7°,  -5°,  0°,  +2°,  +8°,  +8°,  +10°,  +5°,  0°,  -5°.  Plot  this  on 
squared  paper  and  obtain 
the  temperature  line. 

Two  lines  at  right  angles 
are  chosen  as  axes,  Fig.  70. 
Time  in  hours  is  measured 
on  the  horizontal  axis,  with 
one  large  space  to  the  hour. 
The  temperature  in  degrees 
is  measured  vertically  up-        ^^«-  70- -  Timb-Tempebature  Graph 
ward  and  downward  with  one  small  space  to  each  degree.     Thus  for 
9  a.m.  we  count  3  spaces  to  the  right  and  5  spaces  down,  locating  a 
point.     In  this  same  way  we  obtain  a  set  of  points  for  all  of  the  data. 
By  joining  these  points  the  curve  is  obtained. 

From  this  temperature  line  we  may  obtain  various  kinds  of  informa- 
tion, e.g. :    When  was  it  warmest  ?     When  coldest  ?     When  was  the 


s 

r^ 

10 

6 

/ 

\ 

0 

f^ 

/ 

\ 

Ho 

urs 

-5 

6 

7 

8 

y 

\o 

11 

12 

1 

2 

3 

N 

5 

-10 

i-^ 

^ 

^ 

_J 

V,  §  79] 


GRAPHIC  REPRESENTATION 


91 


temperature  changing  most  rapidly?  When  was  the  change  a  rise? 
What  way  would  you  suggest  of  distinguishing  between  a  rising  change 
and  a  falling  change  ? 

79.  The  Graph.  Such  a  diagram  as  that  of  §  78  is  called 
a  graph.  It  furnishes  a  practical  means  of  studying  the 
relations  between  the  two  sets  of  quantities  involved  (time 
of  day  and  temperature  in  this  case).  New  information 
may  be  obtained  from  the  graph. 

EXERCISES 

1.  The  rainfall  in  inches  at  Buffalo,  N.  Y.,  on  the  six  days  following 
April  6,  1908,  was  .46,  .02,  .20,  .14,  0,  .14.     Represent  this  graphically. 

2.  Using,  in  any  city  or  town  famihar  to  you,  any  two  streets 
that  run  at  right  angles  to  each  other  as  axes,  locate  several  points  of 
interest  in  the  city. 

3.  The  following  figures  give  the  temperature  at  intervals  of  three 
hours  of  a  typhoid  fever  patient  in  a  New  York  Hospital,  beginning  at 
11  P.M.  and  running  several  days :  102.2,  102,  101.5,  101.4,  99.6,  101.5, 
101.8, 102,  102,  102.4,  101.3,  101,  99.6,  100.6,  102.7,  102.3,  102.5,  101.5, 
101,  101.7,  99.8,  100.5,  102.4,  102.2,  102.2,  101,  101,  100.4,  99.3,  100, 
101.5,  102.5,  102.2,  101.3,  100.7,  100.4,  100.  Trace  the  rise  and  fall  of 
temperature  by  means  of  a  curve.   This  curve  is  characteristic  of  typhoid. 

4.  The  daily  gain  in  weight  of  a  calf  in  pounds  for  periods  of  one 
hundred  days  is  given  in  the  following  table. 


Age 
in  a  ay  8 

0 

100 

200 

300 

400 

500 

600 

700 

800 

900 

1000 

1100 

1200 

Daily  gain 
in  pounds 

3.2 

2.8 

2.55 

2.3 

2.16 

2 

1.9 

1.8 

1.7 

1.63 

1.57 

1.5 

1.47 

Draw  a  curve  showing  this  information.  Days  may  be  plotted  on 
the  horizontal  scale  and  pounds  on  the  vertical  scale.  When  is  the 
gain  most  rapid? 

5.  The  relation  between  the  age  of  a  calf  and  the  cost  in  dollars 
per  100  pounds  gain  in  weight  is  given  in  the  following  table : 


Age  in  days 

0 

100   200   300 

400 

500 

600 

700 

800 

900 

1000 

1100 

1200 

1300 

1400 

1500 

Cost  per 
100  lb.  Kain 
in  weii?ht 

S3 .40 

3.83  4.25  4.86 

1          1 

5.40 

8.08 

6.75 

7.50 

8.45 

9.56 

10.70 

12.00 

13.35 

14.70 

16.10 

17.85 

Draw  the  graph  for  this  information. 


92 


MATHEMATICS 


[V,  §  79 


6.  The  losses  from  hog  cholera  per  1000  of  hogs  in  the  United 
States  for  the  years  ending  April  1,  1894,  to  1917  inclusive,  are  given 
in  the  following  table. 


Year 

i 

80 

115 

1 

129 

83 

00 

75 

68 

1 
67 

05 

46 

52 

i 

52 

05        03 

47    46 

43 

48 

47 

2  :: 

05         OS 

45    44 

05 

80 

s 

100 

S 

105 

2  '2 

85  60 

2 

44 

Losses 

44 

Show  in  graphic  form  the  annual  losses  from  hog  cholera  throughout 
this  term  of  years. 

Note.  Hog  cholera  serum  came  into  use  about  1907.  Examine 
the  curve  with  this  fact  in  mind  but  remember  also  that  during  the 
years  when  the  curve  was  ascending  little  was  known  regarding  the 
effectiveness  of  hog  cholera  serum,  that  its  use  was  but  imperfectly 
understood,  and  that  during  these  years  the  supply  of  serum  was  far 
short  of  the  demand. 

7.  The  lengths  of  a  thousand  ears  of  corn  were  measured  and 
arranged  in  half -inch  groups  as  follows : 


Length  in  inches    .     .     . 

3.0 

3.. 

4.0 

4.5  5.0 

5.5 

6.0 

6.5 

7.0 

7.5 

8.0 

8.5 

9.0 

9.5 

10.0 

10.5 

11.0 

11.6 

Number  in  each  group 

5 

6 

13 

17    18 

55 

e. 

73 

80 

98 

113 

134 

142 

100 

53 

26 

5 

1 

Plot  these  measurements  and  draw  the  graph,  using  lengths  on  the 
horizontal  scale  and  the  number  of  ears  of  any  given  length  on  the 
vertical  scale. 

From  the  graph  what  would  seem  to  be  an  "average"  length  of 
these  thousand  ears  of  corn?  In  Chapter  X  it  will  be  shown  how  to 
find  this  average  more  definitely. 

8.  The  temperatures  as  observed  on  a  certain  day  in  January, 
1918,  at  Columbus,  Ohio,  were  the  following. 


Midnight  -  7"  F. 

1  A.M.   -    8 

2  A.M.   -    9 

3  A.M.   -10 

4  A.M.   -12 

5  A.M.  —15 

6  A.M.   -16 

7  A.M.   -16 

8  a.m.  -15 


9  A.M.  -  9°  F. 

10  A.M.  0 

11  A.M.  +    9 

12  A.M.  +15 

1  P.M.  +18 

2  P.M.  +20 

3  P.M.  +21 

4  P.M.  +20 


5  P.M.  +16°  F. 

6  P.M.  +12 

7  P.M.  +  8 

8  P.M.  +  2 

9  P.M.   +    1 

10  P.M.  0 

11  P.M.    -    2 

Midnight  -  6 


V,  §  79] 


GRAPHIC  REPRESENTATION 


93 


Draw  the  graph  of  this  information  on  rectangular  coordinate  paper. 
How  long  did  it  take  the  temperature  to  rise  from  its  lowest  to  its  highest 
point  ?  What  was  the  average  rate  in  degrees  per  hour  during  this  time  ? 
When  was  the  temperature  rising  more  rapidly  than  this  average  rate  ? 

9.  Due  to  the  curvature  of  the  earth,  objects  on  the  earth's  surface 
are  visible  only  from  corresponding  elevations  above  the  earth's  sur- 
face.    From  the  approximate  rule* 

Elevation  (in  feet)  =  \  (Distance  in  miles)  2, 
s 

the  following  table  of  corresponding  values  is  obtained. 


Elevation  in 
feet 

1 

5 

10 

20 

30 

40 

60 

100 

160 

200 
17.3 

300 
21.2 

500 
27.4 

1000 
38.7 

5000 
86.6 

10.000 

20.000 

Distance  in 
miles 

1.23 

2.74 

3.87 

5.48 

6.71 

7.76 

8.66 

12.3 

15 

123 

173 

Plot  the  graph  of  these  relations.  Would  there  be  any  advantage 
in  using  a  scale  for  elevations  from  1  to  100  ft.  different  from  that  used 
for  the  other  elevations?     Explain. 

From  the  graph,  how  far  can  a  man  see  if  he  is  6  ft.  tall?  From  the 
rule,  how  far  can  he  see  ? 

10.  Of  100,000  persons  bom  alive  at  the  same  time,  the  following 
table  shows  the  infant  mortality  during  the  months  of  the  first  year. 


Month 

Deaths 

Month 

Deaths 

Month 

Deaths 

1 

4377 

5 

705 

9 

492 

2 

1131 

6 

635 

10 

456 

3 

943 

7 

579 

11 

421 

4 

801 

8 

533 

12 

389 

Draw  the  graph  for  these  data,  plotting  number  of  deaths  on  the 
vertical  scale. 

11.  Draw,  on  the  same  diagram,  graphs  of  the  following  time  table, 
using  distances  on  the  vertical  scale  and  time  on  the  horizontal  scale. 
Mark  the  name  and  the  location  of  each  station  on  the  vertical  scale. 
In  one  case  Columbus  is  the  origin,  in  the  other  Zanesville  is  taken  as 
origin. 

*  For  an  explanation  of  this  rule,  see  §  160. 


94 


MATHEMATICS 


[V,  §  79 


Miles 

Stations 

Time 

Time 

Miles 

Stations 

Time 

Time 

0 

Columbus 

5:55 

7:55 

0 

Zanesville 

5:  55 

7:55 

11 

Reynoldsburg 

6:33 

8:38 

13 

Nashport  .     . 

6:28 

8:28 

15 

Wagram    .     . 

6:40 

8:45 

18 

Blackhand     . 

6:37 

8:37 

18 

Etna     .     .     . 

6:47 

8:50 

21 

Hanover    .     . 

6:44 

8:44 

22 

Kirkersville   . 

6:55 

8:58 

22 

Clay  Lick       . 

6:46 

8:46 

28 

Hebron 

7:05 

9:05 

28 

Newark     .     . 

7:05 

9:05 

37 

Newark 

7:30 

9:30 

28 

Newark     .     . 

7:10 

9:  10 

37 

Newark     .     . 

7:35 

9:35 

37 

Hebron      .     , 

7:36 

9:36 

43 

Clay  Lick       . 

7:55 

9:55 

42 

Kirkersville   . 

7:48 

9:48 

44 

Hanover    .     . 

7:57 

9:57 

47 

Etna     .     .     . 

7:55 

9:55 

47 

Black  Hand  . 

8:04 

10:04 

50 

Wagram    .     . 

8:01 

10:01 

51 

Nashport  . 

8:  12 

10:  12 

54 

Reynoldsburg 

8:08 

10:08 

64 

Zanesville 

8:45 

10:45 

64 

Columbus 

8:45 

10:45 

If  the  railroad  is  single  track,  where  and  at  what  time  should  the 
trains  on  the  above  schedule  pass? 

12.  Draw  a  graph  which  is  slowly  increasing.  One  which  is  rapidly 
increasing.     Draw  slowly  and  rapidly  decreasing  graphs. 

13.  Draw  a  graph  which  is  increasing,  but  which  is  increasing  more 
rapidly  from  point  to  point  along  the  curve.  One  which  is  increasing 
but  less  rapidly  from  point  to  point. 

14.  Draw  a  graph  which  is  decreasing,  but  more  rapidly  decreasing 
from  point  to  point.     Decreasing,  but  less  rapidly. 

Which  of  the  above  cases  would  be  illustrated  by  the  amount  of 
snow  collected  at  each  turn  in  rolling  up  a  snowball? 

Is  the  phrase,  "The  rich  grow  richer,  the  poor,  poorer,"  an  example? 
Would  a  tax  system  based  on  the  rule  that  the  more  property  one  owns 
the  higher  the  rate,  be  an  example? 

Would  the  application  of  fertilizers  to  the  growing  of  crops  be  an 
illustration  of  any  of  these  cases  of  more  rapidly  increasing  graphs? 

15.  The  following  table  shows  the  height  in  inches  of  a  castor  bean 
plant  on  different  days. 


Days      ...... 

0 

20 

40 

60 

80 

100 

120 

140 

150 

160 

Height 

0 

8 

24 

34 

42 

54 

72 

110 

130 

145 

Draw  the  graph  and  from  it  find  the  rate  of  growth, 
rate  of  growth  most  rapid,  slowest  ? 


When  was  the 


V,  §  80]  GRAPHIC   REPRESENTATION  95 

80.  Leveling.  The  art  of  determining  and  representing 
the  inequaUties  in  the  surface  of  any  portion  of  the  earth 
is  called  leveling.  The  principle  is  to  find  the  elevation 
of  each  point  above  some  assumed  base  or  datum. 

If  a  line  of  sight  be  made,  by  any  means,  to  revolve  about 
a  vertical  line  to  which  it  is  perpendicular,  a  horizontal  plane 
is  described.  If  a  rod,  graduated  from  the  bottom  up,  be 
held  at  any  point  on  the  ground,  it  will  be  cut  by  this  hori- 
zontal plane  at  a  distance  above  ground  equal  to  the  height 
of  the  line  of  sight  above  this  point  on  the  ground.  This 
distance  from  the  bottom  of  the  rod,  indicated  by  its  gradua- 
tions, is  called  the  reading  of  the  rod.  If  the  elevation  of 
the  line  of  sight  is  known  with  reference  to  the  assumed 
base  or  datum,  we  can  find,  by  subtracting  the  rod  reading 
from  this  elevation,  the  elevation  of  the  point  on  the  ground. 

The  horizontal  line  of  sight  is  secured  by  means  of  an 
instrument  designed  for  the  purpose  called  a  level. 

For  example,  if  the  rod  reading  at  A,  Fig.  71,  is  2.73' 
and  at  B  it  is  12.15',  and  if  the  line  of  sight  is  100'  above  the 
datum,  then  the  elevation  of  A  above  the  datum  is  97.27' 


Datum  Plane    (100  ^  below  line  of  sight) 


Fig.  71.  —  Line  of  Sight  and  Datum  Plane  in  Leveling 

and  of  B  is  87.85'.  Conversely,  if  the  elevation  of  the  point 
Ls  given,  then  the  elevation  of  the  line  of  sight  above  the  datum 
is  found  by  adding  the  rod  reading  to  the  elevation  of  A. 

Again,  if  the  elevation  of  one  point  above  a  base  or  datum 
is  known,  then  the  elevation  of  a  second  point  may  be  found 


96 


MATHEMATICS 


[V,  §  80 


by  applying  the  difference  in  elevation  of  the  two  points  to 
the  known  elevation.  This  may  be  repeated  at  each  station. 
Ordinarily,  stations  are  50'  or  100'  apart. 

A  diagram  or  graph  of  the  elevations  showing  in  detail 
the  rise  and  fall  of  the  surface  over  which  the  course  passes, 
I.e.  a  vertical  section  of  the  course,  is  called  a  profile  of  it. 
The  profile  is  made  by  plotting  the  elevations  of  the  points 
to  scale  at  the  proper  horizontal  distances  apart  and  con- 
necting these  plotted  points  by  a  line.  Profile  paper  is 
used  for  this  purpose.  It  is  ruled  with  small  divisions  be- 
tween horizontal  lines  and  large  divisions  between  the  vertical 
lines.  The  vertical  distances  are  exaggerated  since  they  are 
small  as  compared  with  the  horizontal  distances. 


EXERCISES 
1.   The  following  level  readings  in  feet  were  taken  at  stations  100  feet 
apart  along  Grafton  Road  : 


station  No. 

0 

19.9 

1 
18.3 

2 
16.2 

3 
15 

4 
16.3 

5 

16 

6 
16.7 

7 
16.2 

8 
14.5 

9 
13.3 

10 
10.4 

11 
10 

12 
13.7 

13 
17 

14 
16.2 

15 
17.4 

16 
17.8 

17 
18 

Elevation 

Make  a  profile  of  these  readings  using  100  ft.  to  the  inch  for  the 
horizontal  scale  and  20  ft.  to  the  inch  vertically. 

If  a  traction  line  were  to  be  laid  on  this  road,  certain  cuts  and  fills 
would  be  made.  Sketch  such  a  proposed  Une  on  the  profile  and  mark 
the  amount  of  cut  or  fill  at  each  station. 

2.  A  farmer  runs  a  level  line  for  a  ditch.  He  first  takes  the  height 
of  a  stump  in  the  vicinity  of  the  ditch  and  then  the  elevation  every 
100  ft.  along  the  line  of  the  ditch,  driving  stakes  at  these  places.  The 
measurements  are  as  follows  above  datum.  Top  of  stump  40  in. ; 
at  stake  no.  1,  60  in. ;  at  no.  2,  etc.,  36  in. ;  48  in. ;  42  in, ;  39  in. ;  36  in. ; 
40  in. ;  30  in. ;  48  in. ;  24  in. ;  27  in. ;  36  in. ;  41  in. ;  28  in. ;  22  in. ; 
12  in.  Make  a  profile  of  these  measurements  using,  as  near  as  is  con- 
venient, the  same  scale  as  in  the  previous  exercise.  Next  establish 
the  grade,  marking  the  amount  to  be  taken  out  at  each  stake  so  that 
the  ditch  will  be  24  in.  deep  near  the  stump  and  slope  gradually  to 
6  in.  deep  at  the  last  stake. 


V,  §  80] 


GRAPHIC   REPRESENTATION 


97 


3.  The  following  elevations  in  feet,  with  the  bench  mark  (B.M.) 
100'  above  datum,  were  taken  at  stations  100'  apart  along  the  route  of 
a  proposed  ditch. 


Sta. 
No. 

Back 
Sight 

Height  op 
Line  of 
Instru- 
ment 

Rod 
Reading 

Elevation 

OF 

Ground 

Fall 

OF 

Drain 

Elevation 

OP 

Drain 

Depth 

OF 

Drain 

B.M. 

— 

— 

— 

100.00 

— 

— 

— 

5.32 

105.32 

— 

— 

— 

— 

— 

0 

— 

— 

9.43 

95.89 

Outlet 

— 

— 

0 

— 

— 

5.82- 

99.50 

Surface 

95.89 

3.61 

1 

— 

— 

5.54 

99.78 

.35 

96.24 

2 

— 

— 

5.02 

100.30 

.35 

3 

— 

— 

4.60 

100.72 

.35 

4 

— 

— 

5.07 

100.25 

.35 

5 

— 

— 

3.10 

102.22 

.35 

— 

8.41 

110.63 

— 

— 

— 

— 

— 

6 

— 

— 

7.84 

102.79 

.90 

7 

— 

— 

7.07 

103.60 

.90 

8 

— 

— 

6.43 

104.20 

.90 

9 

— 

— 

5.63 

105.00 

.90 

10 

— 

— 

5.23 

105.40 

.90 

11 

— 

— 

4.38 

106.25 

.90 

12 

— 

— 

3.93 

106.70 

.90 

13 

— 

— 

3.39 

107.24 

.90 

— 

6.06 

113.30 

— 

— 

— 

— 

— 

14 

— 

— 

5.30 

108.00 

.50 

15 

— 

— 

5.00 

108.30 

.50 

16 

— 

— 

4.70 

108.60 

.50 

17 

— 

— 

4.05 

109.25 

.50 

18 

— 

— 

3.90 

109.40 

.50 

19 

— 

— 

3.57 

109.75 

.50 

20 

— 

— 

3.13 

110.17 

.50 

First.     Make  a  profile  of  the  surface  elevations  at  each  station. 

Second.  Calculate  and  record  the  elevations  of  the  drain  at  each 
station  if  from  stations  0  to  5  the  rise  per  100'  is  .35',  from  stations  5  to 
13  it  is  .90'  per  100',  and  from  stations  13  to  20  it  is  .50'  per  100'. 

Third.     Calculate  and  record  the  depth  of  the  drain  at  each  station. 


98 


MATHEMATICS 


[V,  §  81 


81.  Price  Curves.  Graphs  may  be  constructed  and  used 
for  determining,  without  computation,  costs  of  different 
quantities  of  goods,  interest  on  money,  etc.  The  following 
example  will  illustrate  the  process. 

Example.  If  lard  sells  at  35^  per  pound,  the  relation  between  the 
number  of  pounds  and  the  cost  is  expressed  by  the  equation, 

Cost  =  35  times  the  number  of  pounds,  i.e. 
2/ =  35  X, 

where  x  is  the  number  of  pounds  and  y  is  their  cost.     If  values  are  given 
to  X,  corresponding  values  may  be  found  for  y  as  in  the  following  table. 


If       x  = 

0 

2 

4 

7 

10 

then  y  = 

0 

70 

140 

245 

350 

7   8    9  10  11    Pounds 

FiQ.  72  —  Pounds-Cost  Graph 


On  rectangular  coordinate  paper  select  two  lines  at  right  angles  to 
each  other  as  axes  and  call  their  point  of  intersection,  0,  the  origin, 
Fig.  72.  On  the  vertical  axis,  OY, 
let  a  space  represent  lOjf  and  on  the 
horizontal  axis,  OX,  let  a  space  rep- 
resent one  pound.  Thus  0  pounds 
and  0  cents  corresponds  to  the 
origin;  2  pounds,  70  cents  deter- 
mines a  point  2  spaces  to  the  right 
of  the  origin  and  7  spaces  up.  This 
is  written  (2, 70) .  Likewise  4  pounds, 
140  cents,  i.e.  (4,140)  determines  a  point  4  spaces  to  the  right  and  14 
spaces  up.  Also  7  pounds,  245  cents,  i.e.  (7,245)  and  (10,350)  determine 
points.  These  points  he  on  a  straight  line.  It  is  called  the  price  curve. 
Price  curves  are  straight  lines  in  many  cases. 

By  looking  at  this  graph  or  curve,  the  cost  of  any  number  of  pounds, 
even  of  a  fractional  number  of  pounds,  may  be  determined.  Thus  to 
get  the  cost  of  8  pounds,  find  the  point  where  a  vertical  line  8  spaces 
to  the  right  meets  the  price  curve;  the  point  horizontally  opposite 
this  on  the  axis  OF  is  the  cost.  This  is  28  spaces  from  0  and  hence 
represents  280  cents.     Similarly,  9^  pounds  cost  333  cents. 


V,  §  82]  GRAPHIC  REPRESENTATION  99 

EXERCISES 

1.  Since  the  price  curve  in  the  Example  solved  in  §  81  is  a  straight 
line,  how  many  of  the  points  would  be  essential  to  draw  it?  Should 
these  be  taken  close  together  or  far  apart  in  order  to  get  the  position 
of  the  price  curve  most  exact  ? 

2.  Draw  the  price  curve  for  coffee  at  38^  per  pound.  From  this 
curve  find  the  cost  of  8^  pounds ;  17  pounds ;  24  pounds. 

3.  Draw  the  price  curve  for  sugar  at  ll^jif  per  pound.  By  use  of 
this  curve,  find  the  cost  of  16  pounds ;  42  pounds ;  8^  pounds. 

4.  A  single  railway  ticket  is  to  cost  $2.50.  If  10  tickets  be  pur- 
chased, the  cost  will  be  $2.25  each.  If  50  be  purchased,  the  cost  will 
be  $1.80  each ;  if  100,  the  cost  will  be  $1.50  each ;  and  for  200,  it  will 
be  $1.25  each. 

Draw  a  graph  for  this  information.  Use  it  to  find  the  probable  cost 
per  ticket  for  a  party  of  75.     For  a  party  of  125. 

About  how  many  tickets  must  be  used  to  reduce  the  expense  per  head 
to  $2.00?     To  $1.60? 

5.  A  certain  kind  of  desk  is  quoted  at  $30.00  each.  If  one  dozen 
are  ordered,  they  will  cost  $28.50  each.  For  6  dozens,  the  price  is  $22.50 
each,  and  for  150  the  price  is  $20.00  each.  Draw  a  graph  of  these  data, 
and  from  it  estimate  the  cost  of  a  single  desk  when  36  are  ordered. 
What,  when  100  are  ordered? 

82.   Other  Methods   of   Comparing   Quantities.     So   far 

the  statistical  data  which  we  have  studied  were  exhibited 
on  a  graph  or  curve  drawn  with  reference  to  two  rectangular 
axes.  There  are  various  other  methods  and  devices  for 
representing  quantities.     Some  of  these  are  as  follows. 

(1)  Line  Segments  including  Bar  Diagrams. 

(2)  Area  Charts  such  as 

(a)  Squares  and  Rectangles. 
(6)  Sectors  of  Circles. 

(c)  Dots. 

(d)  Differently  Shaded  Areas,  as  on  maps  of  portions 

of  the  earth's  surface. 


100 


MATHEMATICS 


[V,  §83 


83.  Line  Segments.  The  first  example  given  below  will 
illustrate  a  variation  in  the  use  of  the  graph.  The  second 
example  illustrates  the  use  of  bar  diagrams. 

Example.  1.  According  to  government  regulations,  the  cost  of  do- 
mestic money  orders  is  given  by  the  following  graph. 


I  20 

•Sl5 
"812 
.§10 
^  8- 
6 
8 


Amount  bf  Order  in  Dalian 


2.56  10       20       30       40       60       60  75  100 

Fig-  73.  —  Cost  of  Domestic  Money  Orders 

Example  2.     The  following  table  gives  the  areas  before  1914  of 
some  European  countries  before  the  Great  War,  and  of  some  of  the 


Country     . 

Russia 

Texas 

Germany 

France 

California 

Gt.  Britain 

Ohio 

Sq.  Miles    . 

2.095.616 

265.780 

208,830 

207,054 

158.360 

121.027 

41,060 

states  of  the  United  States.     The  adjacent  figure,  Fig.  74,  represents 
them  by  line  segments  or  so-called  bar  diagrams. 


Russia                J 

Texas 

Germany 

France 

California 

Great  Britain 

:oo( 

Os 

Ohio 

isp 

ac 

'' 

•^ 

'■ 

0  10  20  30 

Fig.  74.  —  Areas  by  Bar  Diagrams 

A  great  many  different  variations  of  this  method  are  in  com- 
mon use.  Thus  the  number  of  fruit  trees  in  various  states 
may  be  represented  by  pictures  of  actual  trees,  whose  heights 
are  proportional  to  the  number  of  trees  in  the  states  they  rep- 
resent. It  should  be  stated  whether  the  given  data  are  repre- 
sented by  the  heights  or  by  the  volumes  of  the  objects  pictured. 


V.  §  83] 


GRAPHIC  REPRESENTATION , 


401. 


EXERCISES     '    •    ,',    '      '-',..'>»..>, 

1.   Parcel  post  rates  in  effect  Jan.  1,  1914,  are  given  in  the  following^ 
table.     Exhibit  this  in  graphic  form. 


Weight  of  Parcel 


Not  over  4  oz 

Over  4  oz.  up  to  1  lb.    .     . 

Over  1  lb.  to  2  lb.     .     .     . 

For  each  pound  over  2  lb. 

up  to  20  lb 


Rates 


1st  and  2d  Zones         3d  Zone 


4th  Zone 


1  cent  per  ounce  or  fraction  thereof 


5ff 
li  additional 


Si 
2i  additional 


7^ 
4:^  additional 


2.    Make  a  diagram  of  segments  showing  the  periods  of  incubation 
of  poultry  as  given  in  the  following  table. 


Name  op  Fowl 

Days 

Name  op 
Fowl 

Days 

Name  of  Fowl 

Days 

Common  Hen  . 
Pheasant     .     . 
Duck,  common 

21 

25 

28 

Pea  Fowl 
Guinea     . 
Goose  .     . 

28 
25 
30 

Partridge     .     . 
Duck,  Barbary 
Turkey  .     .     . 

24 
30 

28 

3.  Show  in  graphic  form  the  production  of  wheat  in  1918  in  the 
United  States  and  in  some  other  leading  wheat-producing  countries 
as  given  in  the  following  table. 


Region 

Bushels 

Acreage 

United  States 

Australia 

Canada    

France 

Germany 

Argentina 

917,000,000 
152,400,000 
210,300,000 
147,150,000 
90,140,000 
187,210,000 

59,100,000 
11,500,000 
17,350,000 
10,260,000 
3,590,000 
17,175,000 

102 


MATHEMATICS 


[V,  §  84 


84.  Area  Charts.  In  example  2,  §  83,  the  table  of  areas 
given  there 'was 'shdwn  on  a  bar  diagram.  That  table  of 
areas  will  now  be  shown  in  various  ways  as  follows. 

A.  By  a  series  of  squares. 

The  sides  of  the  different  squares  are  to  each  other  as  the 
square  roots  of  the  respective  areas.     Thus  the  side  of  the 


Russia 


Texas  Germany  France 

California 


Gt.Britain 


□  Ohio 

u 


Fig.  75.  —  Comparing  Areas  by  Squares 

square  which  represents  the  area  of  Russia  is  about  three 
times  that  for  Texas,  etc.  The  squares  are  then  constructed 
to  a  convenient  scale.  Is  the  comparison  as  effective  in  this 
form  as  in  the  bar  diagram  ?     Why  ? 

B.  By  sectors  of  circles. 

Inspection  shows  the  area  of  Russia  to  be  about  twice  that 
of  all  the  others  in  this  table  combined.     It  should  therefore 


Fig.  76.  —  Comparing  Areas  by  Sectors  of  Circles 


V,  §84] 


GRAPHIC  REPRESENTATION 


103 


occupy  f  of  the  circle.     Texas  is  about  -J  the  remainder,  and 
is  so  represented  on  the  circle. 

C  By  use  of  dot  diagrams. 


Russia 


Texas  Germany  France     Call-      Great       Ohio 

fomia   Britain 


Sxplanation'.  ^^H  =  600,000  Sq.Mi. 
FiQ.  76  a.  —  Dot  Diagram 

D.  By  differently  shaded  areas.  This  is  best  illustrated 
by  reference  to  ordinary  maps  of  these  countries,  made  to 
the  same  circle. 

EXERCISES 

1.  Exhibit  in  graphic  form  the  following  table  of  fertilizing  materials 
and  the  per  cent  of  each  element  carried. 


Material 

%  NlTKOQEN 

%  I*H08PH0Ric  Acid 

%  Potash 

Ground  bone      .... 
Dissolved  bone       .     .     . 
Cottonseed  meal     .     .     . 

Tankage   

Fish  scrap 

Wood  ashes 

3.0 
3.0 
7.0 
6.0 
8.0 

22.0 

15.0 

2.5 

11.0 

7.0 

1.5 

1.5 
5.0 

2.    Make  a  dot  diagram  of  the  world's  wheat  production  for  1914 
given  by  the  following  data. 

The  World 3,362,299,000  bushels. 


United  States   . 
Russia     ... 
France     ... 
Hungary  (proper) 
India  (British) . 


891,017,000  bushels. 
776,960,000  bushels. 
319,667,000  bushels. 
105,237,000  bushels. 
312,032,000  bushels. 


CHAPTER  VI 
GRAPHS  IN  ALGEBRA 

85.  The  Equation  and  its  Graph.  We  have  seen  so  far 
that  various  relations  between  two  quantities,  as  these 
quantities  change,  may  be  represented  by  graphs.  In  some 
cases  we  knew  the  law  of  this  change  as  represented  by 
an  equation.  For  example,  for  lard  at  S5^  per  pound, 
the  law  was  : 

Cost  =  35  times  the  number  of  pounds. 

The  graphs  were  constructed  by  using  two  axes  at  right 
angles  to  each  other.  On  these  axes  we  used  the  number 
scales  for  various  quantities  such  as  cost,  pounds,  time, 
distance,  etc.  In  general  then  we  may  use  y  and  x  to  repre- 
sent these  quantities  and  thus  describe  the  relations  between 
them  in  the  language  of  algebra.  This  leads  very  often  to 
an  equation  in  y  and  x.  Thus  for  lard  at  30^  per  pound, 
the  cost,  y,  of  any  number  of  pounds,  x,  is 

y  =  S5x. 

86.  Rectangular  Coordinates.  It  is  thus  evident  that 
when  the  relation  between  two  quantities  is  expressible  by  an 
equation,  this  relation  may  be  exhibited  by  a  corresponding 
graph.  The  following  considerations  are  commonly  accepted 
as  the  basis  for  such  representation.  Let  us  call  the  hori- 
zontal axis  or  scale,  the  x-axis,  and  the  vertical  axis,  the 
y-axis.  There  being  no  restriction  on  these  scales  as  to 
length,  we  give  x  and  y  each  positive  and  negative  values. 

104 


VI,  §  86] 


GRAPHS  IN  ALGEBRA 


105 


Let  the  two  axes  be  drawn  at  right  angles  to  each  other. 
Their  point  of  intersection  at  0  is  called  the  origin,  Fig.  77. 
They  divide  the  plane  into  four  parts,  called  quadrants, 
named  as  shown  in  the  figure. 

From  the  origin  to  the  right  and  from  the  origin  up  is 
regarded  as  positive  in  direction,  but  to  the  left,  also  down, 
is  negative.  We  can  now  locate  definitely  any  point  in  the 
plane  (of  course  with  reference  to  these  two  fixed  lines). 
Thus  the  notation  (2,  3)  means  the  point  located  by  going 
2  units  to  the  right  on  the  x-axis  and  then  from  this  place 


Second 
Quadrant 


|--1(2,3) 


(-4,-2)^- 

Third 
Quadrant 


First 
Quadrant 


C2,-3) 

Fourth 
Quadrant 


Fig.  77 


3  units  up  parallel  to  the  y-axis,  as  shown  in  Fig.  77.  Like- 
wise, (2,  —3)  means  to  go  2  units  to  the  right  on  the  a:-axis 
and  then  3  units  down  parallel  to  the  2/-axis;  (—4,  —2) 
means  4  units  to  the  left  and  2  units  down.  The  distance 
from  the  vertical  axis  is  commonly  called  the  abscissa  of 
the  point,  and  the  distance  from  the  a:-axis  is  known  as  the 
ordinate  of  the  point.  These  two  together  are  called  the 
coordinates  of  the  point.  The  notation  for  the  coordinates 
of  a  point  is  thus  very  precise,  viz.  written  in  parenthesis  with 
the  x-coordinate  or  abscissa  first  and  separated  from  the  y-coor- 
dinate  by  a  comma. 


106 


MATHEMATICS 


[VI,  §  86 


EXERCISES 

1.  Locate  the  points  (1,  5),  (3,  7),  (5,  2),  (9,  -3),  and  (12,  -6). 
Draw  lines  connecting  these  points  in  the  order  given. 

2.  Draw  the  graph  determined  by  connecting  by  a  smooth  curve 
the  points  (-3,  -2),  (-1,  0),  (0,  1),  (2,  h),  (5,  7),  and  (8,  -11). 

3.  Let  (0,  0)  and  (4,  0)  be  two  vertices  of  an  equilateral  triangle. 
Find  the  coordinates  of  the  third  vertex  in  the  first  quadrant.  What 
is  the  area  of  this  triangle?  What  acute  angles  are  involved  in  this 
triangle  ? 

4.  Find  the  area  of  the  triangle  whose  vertices  are  the  points  (2,  3), 
(5,  6),  (4,  7),  by  dropping  perpendiculars  from  each  vertex  on  the 
y-axia. 

87.  Linear  Equations  and  Their  Graphs.  Every  equation 
of  the  first  degree  represents  a  straight  line.  We  saw  above 
that  many  price  curves  are  straight  lines.  This  is  due  to 
the  fact  that  the  cost  is  equal  to  the  price  of  one  article 
times  the  number  of  articles,  i.e. 


(1) 


y=px, 


where  y  is  the  cost  of  x  articles  at  p  units  each.     But  this 
equation  is  equivalent  to  the  ordinary  proportion, 


(2) 


P     1 


Now  in  Fig.  78,  AE  represents  the  cost,  p,  of  one  article, 
since  from  (1)  y  =  p  when  x  =  l.  BD  represents  the  cost  of 
two  articles,  LC  that  of  x  or  any  number  of  articles. 


O  .1        2  h 

Fig.  78.  —  Linear  Graph 


VI,  §  88] 


GRAPHS  IN  ALGEBRA 


107 


The  triangles  OAE,  OBD,  OCL  are  similar  since 

CL  (or  y)  _  PC  (or  x) 
AE  (or  p)     OA  (or  1) 

But  this  similarity  of  triangles  can  hold  only  when  OL  is  a 
straight  line,  and  the  proportion  (2)  holds  only  when  the 
equation  (1)  is  of  the  first  degree. 

Therefore  an  equation  of  the  first  degree  represents  a  straight 
line. 

State  the  converse  of  this  proposition,  and  show  that  it  is 
true  from  Fig.  78. 

88.   Graph  of  an  Equation.     Let  us  construct  the  graph 

for  the  equation 

y  =  2x+4. 

We  first  give  a  value  to  a:,  as  :c  =  0,  and  find  the  corresponding 
value  of  y.  In  this  case  y  =  4:.  For  x=l,  y  k  Q,  etc.,  as  in 
the  following  table. 


X 

y 

0 

4 

1 

6 

2 

8 

3 

10 

-1 

2 

-2 

0 

-3 

-   2 

-8 

-12 

Fig.  79 


Now  locate,  as  in  Fig.  79,  the  points  whose  coordinates 
are  these  pairs  of  x  and  y  values.  Thus  x  =  Q,  2/  =  4,  i.e. 
(0,4)  locates  the  point  A  on  the  diagram;  (  —  3,  —2)  locates 
B.  It  appears  that  the  graph  is  a  straight  line.  Why  is 
it  a  straight  line? 


108  MATHEMATICS  [VI,  §  88 

EXERCISES 

Find  several  points  on  the  graph  of  each  of  the  following  equations, 
and  draw  the  graph. 

1.  y  =  x,  y-\-x  =  0.  g     ^4-^  =  1 

*    2      "^ 

2.  y+x-2  =  0. 

7.    -^+^=1. 

3.  y  =  x-\-^.  4     6 

4.  2x-Zy  =  Q.  8.    |-?/  =  l- 

5.  a:+?/  +  l=0.  9.    ?/-l=3(x-2). 

10.  What  is  the  degree  in  x  and  y  of  each  of  the  equations  just 
plotted?  What  kind  of  graph  does  each  represent?  What  is  the 
reason  for  calling  these  equations  linear  equations? 

11.  Compare  the  straight  lines  y  =  x  and  y  =  x-\-^.  If  a:  be  given 
the  same  value  in  each  equation,  how  are  the  corresponding  ^/-values 
related?  Try  several  values,  say  x  =  l,  x  =  2,  a:  =  4.  Thus  the  values 
x  =  l,  y  =  l,  fix  a  point  on  the  first  line;  the  values  x  =  \,  ?/  =  4,  fix  a 
point  on  the  second  line.  These  ?/- values  differ  by  3.  Try  other  values 
of  x. 

12.  What  happens  to  the  equation  when  a  straight  line,  say  y  =  x-\-l, 
is  raised  three  units  parallel  to  itself?  Raised  five  units?  Lowered 
four  units  ? 

13.  Plot  y  =  2x.  This  is  the  equation  of  a  straight  line  through 
the  origin,  since  for  x  =  0,yisO.  What  is  the  equation  of  a  straight  line 
parallel  to  y  =  2  x  and  3  units  above  it  ?  How  much  larger  is  every 
y-vahie  ? 

14.  Suppose  that  the  cost  of  setting  the  type  in  a  job  of  printing 
is  $1.50,  and  the  cost  of  paper  and  printing  is  ^^  per  copy.  Find  the 
cost,  y,  of  any  number  of  copies,  x  (i.e.  y  =  total  cost  and  x  =  any  number 
of  copies). 

Draw  the  price  curve  from  the  cost  equation. 

[Hint.  In  plotting  the  equation,  let  one  vertical  small  space  equal 
lOff  and  one  horizontal  small  space  equal  10  copies.] 

15.  The  cost  of  setting  the  type  for  an  order  of  printing  is  $1.00. 
The  cost  of  material  and  printing  is  ^^  for  each  copy.  What  is  the  cost 
for  any  number  of  copies  ?  Draw  the  graph.  Estimate  from  the  graph 
the  price  of  150  impressions.     Of  500  copies. 


VI,  §  89]  GRAPHS  IN  ALGEBRA  109 

16.  For  the  job  of  printing  in  Ex.  15,  another  firm  offers  to  disre- 
gard the  initial  cost  of  typesetting,  and  to  print  the  copies  for  ^^  each. 
Find  the  cost  for  any  number  of  copies.  Draw  the  graph  on  the  same 
diagram  as  for  the  previous  example. 

17.  Of  the  two  firms  in  the  Exs.  15-16,  which  one  will  most  cheaply 
furnish  100  copies?  Which  firm  will  give  the  greater  number  of  copies 
for  $1.25?  For  $2.00?  How  many  copies  will  be  printed  just  as 
cheaply  by  one  firm  as  by  the  other  ?     What  will  they  cost  ? 

18.  The  initiation  fee  in  a  certain  society  is  $5.00  and  the  annual 
dues  are  $3.00.  What  is  the  cost  of  membership  for  any  number  of 
years?  Draw  the  graph.  Estimate  from  this  graph  the  cost  of  mem- 
bership for  5  years.     For  12  years. 

19.  The  cost  of  mimeographing  is  quoted  by  a  firm  as  follows : 
25  copies  of  1  page,  50ff;  50  copies  of  1  page,  57^;  100  copies,  70^; 
200  copies,  95  ff;  300  copies,  120^;  400  copies,  145^;  500  copies,  170^ 
Plot  this  data,  and  draw  the  price  curve.     What  is  the  graph? 

20.  Draw  a  graph  showing  the  relation  between  the  Fahrenheit 
and  Centigrade  thermometers. 

Plot  the  Fahrenheit  scale  as  the  vertical  axis  and  the  Centigrade 
scale  as  the  horizontal  axis.  The  Fahrenheit  scale  is  marked  32°  at 
the  freezing  point,  while  0°  is  the  freezing  point  on  the  Centigrade  scale. 
Hence  32°  must  be  subtracted  from  Fahrenheit  readings  before  com- 
paring with  the  Centigrade  reading.  Find  the  difference  between  the 
readings  for  freezing  and  for  boiling  on  each  scale.  What  is  the  ratio 
between  these  two  differences?  Establish  the  algebraic  relation,  and 
plot  the  graph  from  it.  Ans.     f'°  =  9/5  C° +32. 

21.  Calculate  algebraically  the  Centigrade  reading  for  41°  F., 
0°  F.,  -20°  F.,  +98°  F.,  +68°  F.  Do  your  answers  correspond  to 
the  graphical  results? 

In  the  same  way  calculate,  graphically  and  algebraically,  the  Fahren- 
heit readings  for  the  following  Centigrade  readings :  14°  C,  25°  C, 
20°  C,  -10°  C. 

89.   Two  Points  Determine  the  Graph  of  a  Straight  Line. 

We  have  seen  that  every  equation  of  the  first  degree  in  two 
variables  represents  a  straight  line.  Since  a  straight  line 
is  fixed  by  any  two  points  on  it,  to  find  the  graph  for  an 


110 


MATHEMATICS 


[VI,  §  89 


equation  of  the  first  degree,  it  is  only  necessary  to  find  two 
points  on  the  graph  and  draw  the  straight  fine  through  them. 

Example.     Plot  the  graph  of  the  equation  3  a:— 4i/  +  12  =  0. 
For  x  =  2,y  is  4^,  and  for  x  =  —  1,  ?/  is  2^.     Locate  the  points  (2,  4^), 
(  —  1,  2^),  and  draw  the  straight  Hne  as  in  Fig.  80. 

90.  Intercepts.  Since  x  =  0  for  every  point  on  the  y-axis, 
to  find  where  any  curve  crosses  the  y-Sixis,  we  put  x  =  0,  and 
find  y.  Similarly  to  find  the  x-inter- 
cept,  put  y  =  0  and  find  x.  In  the 
example  just  given,  viz.  3  a:  — 4 1/ 
+  12  =  0,  if  x  =  0,  y  is  3,  and  when 
2/  =  0,  X  is  —4.  It  is  seen  in  Fig.  80 
that  the  straight  line  passes  through 
the  points  (—4,  0)  and  (0,  3)  on 
the  axes.  Such  distances  cut  off  on  the  axes  by  a  curve  or 
by  a  straight  line  are  called  the  intercepts. 

91.  Simultaneous    Solution   of    Equations.     In    solving 
simultaneous  equations,  for  example,  the  pair 

2x-y=\, 
3x+2  2/=12, 

we  find  the  pairs  of  values  of  x  and  y  which  satisfy  both 
equations.  To  solve  the  given  pair,  multiply  the  first  one 
by  2  and  add  the  result  to  the  second 
equation.      This  gives 

7x=14,  or  x  =  2. 

In  either  equation,  if  x  =  2,  y  =  3.  Thus 
(2,  3)  is  the  set  of  values  that  satisfies  both 
equations. 

Let  us  now  see  what  this  means  with 
reference  to  the  graphs  of  these  equations, 
finding  their  intercepts,  as  in  Fig.  81. 


Fig.  81 


Plot  them  by 
Each  line  passes 


VI,  §  91] 


GRAPHS  IN  ALGEBRA 


111 


through  the  point  (2,  3).  Hence  the  coordinates  for  the 
point  of  intersection  of  two  graphs  is  the  set  of  values  ob- 
tained by  solving  them  simultaneously. 


EXERCISES 

Draw  the  graphs  for  each  of  the  following  pairs  of  equations.  Find 
their  solution  from  this  diagram,  and  determine  whether  this  agrees 
with  their  algebraic  solution. 


1. 


2x-3y  =  25, 
x+y  =  5. 

3.  \ 

5x+2y  =  S, 
,2x-Sy=- 

2x-3y  =  &, 

12. 

5. 

5x+6  2/  =  7, 
2x-4!/=-4. 

4. 

J+I=^- 

6. 

3-4a:  =  y, 
_8x-\-2y  =  Q. 

7.  The  equations  of  the  sides  of  a  triangle  are 

x-y  +  l=0,  2x+2/  =  13,  x+5y  =  n. 

Plot  these  equations  and  find  each  of  the  vertices  of  the  triangle. 

8.  Find  the  length  of  each  of  the  sides  of  the  triangle  of  Ex.  7. 

[Hint.  Use  each  side  of  the  triangle  as  the  hypotenuse  of  certain 
right  triangles  formed  by  drawing  lines  through  the  vertices  parallel 
to  the  axes.] 

9.  A  rectangular  field  is  32  rods  longer  than  it  is  wide.     The  length 
of  the  fence  around  it  is  308  rods.     Find  the  dimensions  of  the  field. 

10.  A  steamer  makes  6  mi.  an  hour  against  the  current  and  19^ 
miles  an  hour  with  the  current.  What  is  the  rate  of  the  current  and 
what  is  the  steamer's  rate  in  still  water? 

11.  The  difference  between  two  sides  of  a  rectangular  wheatfield 
is  30  rods.  A  farmer  cuts  a  strip  5  rods  wide  around  the  field  and 
finds  the  area  of  this  strip  to  be  7^  acres.  Find  the  dimensions  of  the 
field. 

12.  Given  two  boys  A  and  B,  a  30-lb.  weight,  and  a  teeter  board. 
They  find  they  will  balance  when  B  is  6  ft.,  and  A  5  ft.,  from  the  ful- 
crum ;  but  if  B  places  the  30-lb.  weight  on  the  board  beside  him,  they 
will  balance  when  B  is  4  ft.,  and  A  5  ft.,  from  the  fulcrum.  Find  the 
weight  of  each  boy,  using  the  law  of  levers ;  that  the  weights  balanced 
are  inversely  proportional  to  their  respective  distances  from  the  fulcrum. 


112  MATHEMATICS  [VI,  §  92 

92.   Sine,   Cosine,  Tangent   of  an   Obtuse  Angle.     Let 
XOB,  Fig.  82,  be  an  obtuse  angle.     From 
a  point  on  OB  drop  a  perpendicular,  EC     \js 
on  XO  produced.     OC  is   then   negative,       |\^ 
§  76,  and  OB  is  regarded  as  positive.     The       !        X~^ 
sine,    cosine,    tangent    of    XOB   are    now      ^  ^ 

.    _       -   ,         ,       „  T,       .  .  Fig.  82.  — Obtuse 

denned  by  the  following  ratios:  Angle 

sin  XOB=  ^^,  and  is  +, 
OB'  ' 

cos  XOB=  —  ,  and  is  — , 

CB 

tan  XOB  =-^^,  and  is  —. 

OC' 

It  follows  that  the  sine  of  the  obtuse  angle  XOB  is  identically 
equal  to  the  sine  of  its  supplementary  acute  angle  COB ;  and 
that  the  cosine  and  the  tangent  of  the  obtuse  angle  XOB  are 
numerically  equal,  respectively,  to  the  cosine  and  the  tangent 
of  its  supplementary  acute  angle  COB,  but  opposite  in  sign. 

Hence,  if  A  is  any  obtuse  angle  and  180°  — A  its  supple- 
ment, we  have 

(1)  sin  {im-A)  =  sinA. 

(2)  cos  {1^0- A)  =- cos  A. 

(3)  tan  (180- A)  = -fan  A. 

Examples. 

sin  150°  =  sin  (180°  - 150°)  =  sin  30°  =  +  .5 

cos  150°=  -cos  (180° -150°)  =  -cos  30°=  -.8660 

tan  150°  =  -tan  (180° - 150°)  =  -tan  30°  =  - .5774 

sin  110°=  sin  (180° -110°)  =sin  70°  =  .9397 

cos  110°=  -cos  (180° -110°)  =  -cos  70°=  -.3420 

tan  110°=  -tan  (180° - 110°)  =  -tan  70°=  -2.747 


VI,  §  93]  GRAPHS  IN  ALGEBRA  113 

93.  Reading  of  Tables  for  Obtuse  Angles.  To  find 
from  the  table  the  value  of  the  sine,  cosine,  or  tangent  of 
an  obtuse  angle,  find  the  value  of  the  same  function  of  the 
supplement  of  the  angle,  and  for  the  cosine  and  the  tangent 
prefix  the  negative  sign. 

Examples,  sin  1 12°  10'  =  +sm  67°  50'  =  +  .9261 
cos  112°  10'=  -cos  67°  50'=  -.3773 
tan  112°  10'=  -tan  67°  50'=  -2.4545 

Conversely,  if  cos  A  or  tan  A  is  negative,  find  from  the 
table  the  angle  whose  cosine  or  tangent  is  respectively  this 
same  positive  value.  Then  subtract  this  angle  from  180°. 
The  result  is  A. 

Examples.     Find  A,  if  cos  A  =  —.8192 

Look  for  +.8192  in  the  column  of  cosines.  This  corresponds  to  35°. 
Then  .1  =  180 -35°  =  145°. 

Again,  if  tan  A  =  —.5774,  find  A. 

Look  for  +.5774  in  the  column  of  tangents.  This  corresponds  to 
30°.     Then  A  =  180°  -30°  =  150°. 

If  sin  A  has  a  given  positive  value,  there  are  two  angles 
which  have  this  sine,  one  an  acute  angle  and  the  other  its 
supplement. 

EXERCISES 

1.  Given  sin  A  =  3/5,  where  A  is  an  obtuse  angle.    Find  cos  A ,  tan  A . 

2.  Given  tan  A  =  —5/12,  where  A  is  an  obtuse  angle.     Find  sin  A, 
cos  A. 

3.  A  flagstaff  80  feet  high,  on  a  horizontal  plane,  casts  a  shadow  110 
feet  long.     Find  the  angle  made  by  the  sun's  rays  with  the  horizontal. 

Find  the  value  of  the  following  expressions  from  a  table  of  natural 
functions. 

4.  sin  131°  10'  6.   tan  120°  8.   tan  95°  15' 

5.  cos  150°  20'  7.   cos  95°  15'  9.   sin  95°  15' 
Find  A  in  each  of  the  following  expressions. 

10.  sin  ^  =  .4746  12.   cos  A  =  -.4746         14.   tan  A  =  -.4125 

11.  cos  A  =  -.8090         13.   tan  A  =  -1.7321      15.   cos  A  =  -.5819 


114 


MATHEMATICS 


IVI,  §  94 


94.  Grade,  Slope.  In  an  earlier  section,  §  4,  we  spoke  of 
the  pitch  of  a  roof.  It  was  defined  as  the  rise  of  the  roof 
divided  by  the  width  of  the  building.  Applied  to  a  hill, 
we  saw  that  the  rise  of  the  hill  divided  by  its  run  is  its  slope 
or  its  grade.  Mention  was  made  also  of  the  fact  that 
many  grades,  particularly  in  railroad  construction,  are 
spoken  of  in  terms  of  per  cent ;  for  example,  a  5%  grade  is 
one  that  rises  5'  in  100'. 

A  little  later,  §  36,  this  notion  was  made  more  precise  in 
studying  the  right  triangle.  The  rise  of  the  hypotenuse  of 
a  right  triangle  divided  by  its  run,  i.e.  the  altitude  divided  by 
the  base,  was  used  as  a  measure  of  the  base  angle,  and  was 
called  the  tangent  of  this  angle. 

The  same  notion  of  slope  will  be  used  in  what  follows  for 
any  straight  line. 

95.  Slope  of  a  Straight  Line.  Let  a  straight  line  pass 
through  two  points  P={xi,  yi)  and  Q  =  {x2, 2/2),  Fig.  83.  Drop 
perpendiculars  parallel  to  the  axes  as  shown.     Then 


slope  of  PQ 


yi 

PA         X2  —  Xt 


Thus  the  slope  of  a  line  between  two  points  is  equal  to  the 
difference  of  the  y-coordinates  of  the  points  divided  by  the 
difference  of  their  x-coordinates,  subtracted  in  the  same  order. 

In  Fig.  83  a,  the  slope  of  PQ  is  positive,  since  it  is  the  rise 
divided  by  the  run,  in  moving  along  the  line  to  the  right. 
Conversely,  a  rise  divided  by  a  run  means  a  positive  slope. 

Y 


Positive  Slope 


Fig.  83  6.  —  Negative  Slope 


VI,  §  95]  GRAPHS  IN  ALGEBRA  115 

This  corresponds  to  the  tangent  of  the  acute  angle  XKP 
which  this  Hne  makes  with  the  x-axis  on  the  right  side  of 
the  hne. 

But,  in  Fig.  83  6,  the  slope  of  PQ  is  negative,  since  it  is  the 
fall  divided  by  the  run,  in  moving  to  the  right  along  the 
line.  Conversely,  fall  divided  by  run  means  a  negative 
slone.  This  corresponds  to  the  tangent  of  the  obtuse  angle 
XKP  which  this  line  makes  with  the  x-axis  on  the  right  side 
of  the  line,  §  92. 

Example  1.  Find  the  slope  of  the  line  through  the  two  points 
P  =  (3,  4)  and  Q  =  (8,  6).  Find  also  the  angle  which  this  line  makes 
with  the  X-axis,  as  angle  XKP,  Fig.  83  a. 

Solution  :  Taking  the  two  points  in  the  order  P,  Q,  we  find 

SlopeofPQ  =  |^  =  ^  =  +.4, 
or,  if  the  points  are  taken  in  the  order  Q,  P,  we  have 

8-3      ^    ' 

so  that  the  answer  is  the  same  in  either  case.     Hence 

tan  XKP  =  A, 
whence 

Z  XKP  =  21°  48'. 

Example  2.  Find  the  slope  of  the  Une  through  the  two  points 
P  =  (3,  7)  and  Q  =  (10,  2).  Find  also  the  angle  which  this  line  makes 
with  the  X-axis  on  the  right  side  of  the  line. 

Solution  : 

Slope  of  PQ  =  ^^  =  -  I  =  -  -7143, 

or 

tan  XKP  =  - .7143,     and     Z  XKP  =  180° -31°  12'  =  148°  48'. 

EXERCISES 

1.   Construct  a  line  through  (0,  0)  whose  slope  is  3/4. 

[Hint.  A  slope  +3/4  means  a  rise  of  3  and  a  run  of  4.  Therefore 
begin  at  (0,  0),  rise  3  units  and  run  4  units  to  the  right.  Connect  the 
final  point  with  (0,  0).     The  resulting  line  will  have  the  slope  3/4.] 


116  MATHEMATICS  [VI,  §  95 

2.  Construct  a  line  through  (0,  0)  whose  slope  is  —3/4. 

[Hint.  A  negative  slope  means  a  fall  and  a  run.  Therefore,  fall 
3  units  from  (0,  0)  and  run  4  units  to  the  right.] 

3.  Construct  a  slope  of  5/7  from  the  point  (1,  2).  Also  one  of 
—5/7  from  the  same  point. 

4.  Compare  the  slopes  3/4  and  —4/3  from  the  same  point  (2,  2). 
Do  you  recognize  the  angle  between  these  slopes? 

5.  Compare,  as  in  exercise  4,  the  slopes  —2/3  and  +3/2,  from 
the  same  point  (3,  4). 

For  each  of  the  following  pairs  of  points : 

(a)  Plot  the  points. 

(fe)  Draw  the  straight  line  through  them. 

(c)   Find  the  slope  of  the  straight  line. 

{d)  Find  the  angle  the  line  makes  with  the  x-axis. 

6.  (1,  2)  and  (3,  3).  9.    (-3,  2)  and  (2,  -4). 

7.  (1,  2)  and  (3,  -1).  10.    (-3,  -4)  and  (-1,  -1). 

8.  (3,  2)  and  (-3,  -4).  11.    (-3,  -4)  and  (-5,  1). 

96.  To  Find  the  Equation  of  a  Straight  Line.  Heretofore 
we  have  had  certain  equations  given  and  our  problem  was 
to  find  the  graphs  of  these  equations.  Our  problem  now  is 
to  find  the  equation  when  the  graph  is  given.  The  problem 
here  is  to  connect  in  some  algebraic  way  the  relation  between 
the  x-distance  and  the  ^/-distance  of  a  point  which  will 
hold  for  all  points  on  the  line. 

For  example,  if  a  point  is  located  anywhere  on  the  x-axis 

the  one  thing  we  can  say  of  this  point  is  that  its  i/-coordinate 

is  always   zero.     The  concise  algebraic  statement  of  this 

fact  is  the  equation 

y  =  0. 

Hence  this  is  the  equation  of  the  x-axis,  since  it  is  the  one 
statement  that  is  true  for  any  and  all  points  on  the  a;-axis, 
and  for  no  other  point. 

What  is  the  equation  of  the  2/-axis  ? 


VI,  §  97] 


GRAPHS  IN  ALGEBRA 


117 


f(x,y) 


Again,  let  us  find  the  equation  of  the  Hne  which  bisects 
the  first  and  third  quadrants  and  passes  through  the  origin. 
Let  (x,  y),  Fig.  84,  be  any  point  on  this  Hne. 
The  problem  is  to  find  some  relation  be- 
tween the  ^/-distance  and  the  x-distance 
which  will  be  true  for  all  points  on  this  line. 
Since  POA  =  45°,  it  follows  at  once  that 
PA  or  2/  is  equal  to  OA  or  x.     This  fact 
is  stated  algebraically  by  the  equation 
y=x,     or     x-y  =  0. 


^^ 


O  A 

y  =  X 

Fig.  84 


97.  Line  through  Two  Points.  Line  through  One  Point 
in  a  Given  Direction.  A  straight  line  is  fixed  definitely  in 
position  if  it  passes  through  two  fixed  points,  or  if  it  passes 
through  one  fixed  point  in  a  given  direction.  The  equation 
of  the  line  in  either  case  may  be  found  by  using  the  slope 
of  the  line.  In  one  case  the  slope  between  the  two  given 
points  may  be  found,  in  the  other  case  it 
is  given.  The  method  will  be  shown  by 
some  particular  examples. 


Example  1 .  Find  the  equation  of  the  straight 
line  which  passes  through  the  point  (3,  2)  and  has 
the  slope  2. 

Solution:  Plot  the  point  (3,  2),  Fig.  85. 
Draw  a  line  through  this  point  with  the  slope  2. 
Let  P{x,  y)  be  any  point  on  the  line.     The  slope  from  A  to  P  is 

BP 


Fig.  85 


But 

BP=PD-AC=y 

It  follows  that 


-2,     and 


AB  =  0D-0C  =  x-3. 


=  2. 


Hence  this  is  the  equation  of  the  line.     Clearing  of  fractions,  we  find 

y  =  2x-4:. 


118 


MATHEMATICS 


[VI,  §  97 


Attention  is  called  to  the  form  of  this  last  equation,  in  which  the 
slope  of  the  line  appears  as  the  coefficient  of  x. 

Example  2.     Find  the  equation  of  the  straight  line  through  (1,  2) 
and  (5,  4). 

Solution:   Plot   these  points  and  draw  the  line  through  them. 
Now  take  any  point  R  on  the  line  with 
coordinates  {x,  y),  Fig.  86.     Then 

slope  P  to  Q  =  slope  Q  to  R, 


R{x,y) 


or 


or 


4-2_i/-4 
5-1      a:-5' 

l_y-4 
2     x-b' 


from  which 


y 


x+|. 


What  is  the  slope  of  the  straight  line?  Does  the  equation  of  the 
straight  line  show  this?     Note  the  form  of  this  equation. 

EXERCISES 

1.  Find  the  equations  of  the  straight  lines  passing  through  the 
following  points  and  having  the  given  slopes. 

(a)  Through  (3,  4)  with  the  slope  2/3. 

{h)  Through  (3,  4)  with  the  slope  -2/3. 

(c)  Through  (-3,  4)  with  the  slope  6/5. 

id)  Through  (2,  -3)  with  the  slope  -3. 

2.  What  is  the  equation  of  a  line  parallel  to  the  x-axis  and  four 
units  above  it  ?     Four  units  below  it  ? 

What  is  the  equation  of  a  line  parallel  to  the  ?/-axis  and  twenty  units 
to  the  right  of  it  ?     Parallel  to  the  ^/-axis  and  three  units  to  the  left  ? 

3.  What  is  the  slope  of  a  straight  line  bisecting  the  first  quadrant? 
Does  this  same  line  (extended)  bisect  the  third  quadrant?     Why? 

Find  the  equation  of  the  straight  line  bisecting  the  second  and 
fourth  quadrants.     What  is  its  slope? 

4.  Find  the  equation  of  a  line  through  each  of  the  following  pairs 
of  points.  Reduce  each  equation  to  the  form  showing  its  slope.  Find 
the  x-intercept  and  the  ^/-intercept  in  each  case. 

(a)  (3,  4)  and  (-2,  2).  {d)  (2,  4)  and  (1,  -1). 

(6)   (3,  2)  and  (5,  6).  (e)    (-15,  -3)  and  (8/3,  -7/9). 

(c)    (-6,  1)  and  (-1,  -5).    (/)    (-1,  0)  and  (4,  -2). 


VI,  §  98]  GRAPHS  IN  ALGEBRA  119 

5.  Write  the  equation  of  each  of  the  straight  Unes  described  below. 

(a)  A  line  whose  x-intercept  is  4  and  whose  y-intercept  is  7,  i.e. 

a  line  through  (4,  0)  and  (0,  7). 

(b)  A  line  whose  x-intercept  is  —3  and  whose  slope  is  4/5. 

(c)  A  line  whose  y-intercept  is  8/3  and  whose  slope  is  —2. 

6.  Find  the  equation  of  a  Une  through  the  origin  (0,  0)  and  the 
point  (5,  3).  Of  course  there  are  no  intercepts  in  this  case.  How  does 
this  fact  appear  in  the  equation? 

7.  Find  the  equation  of  a  straight  line  whose  x-intercept  is  twice 
as  great  as  its  2/-intercept  and  which  also  passes  through  the  point 
(-2,  3). 

[Hint.     What  is  the  slope  of  the  line  ?] 

8.  Find  the  equation  of  the  straight  hne  which  makes  equal  inter- 
cepts on  the  axes  and  which  passes  through  the  point  (2,  —4). 

9.  Find  the  equation  of  the  straight  line  which  makes  an  angle  of 
60°  with  the  x-axis  and  which  passes  through  the  point  (2,  —3). 

[Hint.     What  function  of  the  angle  is  the  slope  ?] 

10.  A  ditch  rises  1  ft.  in  50  ft.  Draw  a  diagram  of  it  and  find  its 
equation. 

11.  Find  all  the  points  for  which  x  =  2,  i.e.  draw  the  Une  whose 
equation  is  a;  =  2.     Do  the  same  for  x=  —2,  0,  1,  3,  —3. 

12.  Draw  the  hues  y  =  0,  1,  —1,  2,  3,  —4,  respectively. 

98.  Parallel  Lines.  Draw  two  non-vertical  parallel  lines 
as  shown  in  Fig.  87.  Select  any  two  points  on  each  line. 
The  slopes  are,  respectively, 

^   and  ^. 

The  triangles  having  these  sides  are 
similar.     Hence  T' 

ai  _  fl2 

61       62'  ^^-  ^^- — Parallel 

and  the  two  slopes  are  equal.     There- 
fore, if  two  non-vertical  lines  are  parallel,  their  slopes  are 
equal.     State  and  prove  the  converse. 


120 


MATHEMATICS 


[VI,  §  99 


99.   Perpendicular  Lines.     Draw  two   non-vertical,   per- 
pendicular lines,  Fig.  88. 
The  slope  of  the  one  line  is 

mi  = -'     (Why  negative  ?) 

Pi 

The  slope  of  the  other  line  is 

02 

But,  as  the  triangles  are  similar. 


a2      hi 

1 

FiQ.  88 

•  —  Perpendicu- 

hi      ai 

ai 

LAR  Lines 

__  J_ 

mi 

i.e. 


Hence,  if  two  non-vertical  lines  are  perpendicular  to  each 
other,  the  slope  of  one  is  the  negative  reciprocal  of  the  slope  of 
the  other. 

EXERCISES 

1.  A  line  passes  through  the  points  (2,  5)  and  (6,  —1).  What  is 
its  slope  ?     What  is  the  slope  of  a  line  perpendicular  to  it  ? 

Ans.     -3/2;  2/3. 

2.  What  is  the  slope  of  the  line  whose  equation  is 

Is  the  point  (4,  5)  on  this  line  ?  What  is  the  slope  of  a  line  perpen- 
dicular to  it  ? 

3.  Discuss  the  slopes  of  the  following  equations. 

(a)  2/  =  2a;+3,  (c)   ^=-ia;+5, 

(6)  2/  =  2a:+4,  {d)  2/=-|x+6. 

4.  Find  the  equation  of  a  line  through  the  point  (1,  —2)  parallel 
t0  2/  =  2x+3. 

[Hint.  From  the  given  equation,  what  is  the  slope  ?  Having  found 
the  slope,  you  can  find  the  equation  as  required.] 


VI,  §  100] 


GRAPHS  IN  ALGEBRA 


121 


MUes 

V 

3 

/ 

2 

/ 

1 

/ 

Hour* 

i 

1      2 

3     X 

1 

5.  Find  the  slope  and  the  intercepts  on  the  axes  for  each  of  the 
following  lines,  and  draw  the  lines. 

{a)  3x-4y  =  12.  {c)x+y  =  5.  ^  +  ^=1 

(6)   5x-{-2y^l0.  (d)  Qx+Sy  =  2.  ^^^2     3 

6.  Give  the  slopes  of  lines  perpendicular  to  the  lines  whose  equations 
are  given  in  Ex.  5. 

100.   Motion  Problems  Involving  Time,  Rate,  Distance. 

Many  problems  involving  the  element  of  time  may  be  solved 
graphically.  If  a  man  walks  6  miles  in  2  hours  at  a  steady 
rate,  he  walks  at  the  rate  of  3  mi.  per  hour. 
This  may  be  represented  on  a  diagram. 
Fig.  89,  by  plotting  time  (in  hours)  on  the 
horizontal  axis  and  distance  (in  miles)  on 
the  vertical  axis  according  to  the  relation 
distance  =  3  times  the  number  of  hours 
traveled, 
or,  what  is  the  same  thing, 

2/  =  3x, 

if  y  means  distance  traveled  and  x  means  hours  traveled. 
This  gives  the  straight  line  OP,  which  represents  geomet- 
rically the  uniform  motion  of  the  man. 

Example.  An  automobile  starts  from  Columbus,  going  30  miles 
per  hour.  Two  hours  later  a  motorcyclist  starts  in  pursuit  at  the  rate 
of  54  miles  per  hour.  If  they  continue  at 
these  rates,  when  and  how  far  from  the 
starting  point  will  the  motorcycle  overtake 
the  automobile  ? 

In  the  diagram.  Fig.  90,  let  each  large 
horizontal  space  represent  1  hr.,  and  each 
large  vertical  space  30  mi.  Then  in  1  hr. 
the  auto  goes  30  mi.,  in  2  hr.  60  mi.,  etc.  The 
points  representing  this,  i.e.  (1  hr.  30  mi.), 
(2  hr.  60  mi.),  etc.,  lie  on  a  straight  line  OP. 

The  motorcyclist  starts  at  zero  miles  after  2  hours.    This  is  represented 
by  the  point  M.     He  goes  54  mi.  in  1  hr.     Hence  his  motion  is  repre- 


FiG.  89.  — Time- Dis- 
tance Graph 


Y 

MUes 

,/ 

150 

■ 

/ 

120 

■ 

r 

90 

60 
30 

A 

/ 

---y{/ 

/\     Vm 

Hours 

/ 

1/2     3 

4      5     X 

Fig.  90 


122  MATHEMATICS  [VI,  §  100 

sented  by  the  graph  MP.  The  point  where  these  two  graphs  intersect 
shows  the  place  and  time  of  meeting.  From  the  diagram  the  time  is 
4|  hr.  and  the  distance  is  135  mi.  Solve  the  problem  algebraically  and 
thus  check  the  accuracy  of  the  graphical  solution. 

Of  course,  the  graphical  solution  is  not  always  exact. 
In  most  cases  it  is  only  approximate,  but  in  any  case  it 
serves  as  a  check  and  a  guide  to  the  algebraic  solution. 
However,  all  measurements  are  only  approximations,  — 
accurate  only  to  a  certain  degree.  In  the  preceding  prob- 
lem, the  rate  of  30  mi.  per  hour  is  only  an  approximation- 
Thus  our  graphical,  approximate  solutions  are  frequently 
as  accurate  and  rehable  as  the  data  from  which  these  solu- 
tions are  made. 

For  example,  in  transferring  degrees  Fahrenheit  to  degrees 
Centigrade  by  the  graphical  process  the  approximate  answer 
is  often  desirable.  In  fact  very  few  thermometers  except 
those  made  especially  for  scientific  purposes  are  really 
accurate.  Few  thermometers  give  the  correct  temperature 
to  within  half  a  degree.  In  no  case  should  an  answer  be 
expressed  more  accurately  than  the  circumstances  warrant. 

EXERCISES 

[Solve  the  following  problems  graphically.] 

1.  A  starts  for  a  town  12  mi.  distant,  walking  at  the  rate  of  3  mi. 
per  hour.  An  hour  and  a  half  later  B  starts  for  the  same  place,  driving 
at  the  rate  of  7^  mi.  per  hour.  When  does  B  overtake  A?  Where  is 
A  when  B  reaches  the  town  ? 

2.  In  a  mile  race,  A  runs  6  yd.  per  second  and  B  5  yd.  per  second. 
If  B  starts  250  yd.  ahead  of  A,  who  will  win  the  race  and  how  far  ahead 
is  the  winner  at  the  finish  ? 

[Hint.  Plot  one  small  vertical  space  equal  to  20  yd.  and  one  small 
horizontal  space  equal  to  5  sec] 

3.  At  6  A.M.  a  freight  train  leaves  a  station  0  going  north  at  the 
rate  of  30  mi.  per  hour.     At  9  a.m.  an  express  train  is  due  at  this  same 


VI,  §  100]  GRAPHS  IN  ALGEBRA  123 

station  coming  from  the  opposite  direction  at  60  mi.  per  hour.     When 
and  where  shall  these  trains  be  ordered  to  pass  ? 

4.  How  long  will  it  take  a  steamboat,  going  at  the  rate  of  8  mi.  per 
hour,  to  overtake  a  motor  boat  10  mi.  ofif,  going  directly  away  at  the 
rate  of  5  mi.  per  hour? 

5.  A  train  traveling  30  mi.  per  hour  is  followed  3  hr.  after  its  start 
by  a  train  making  50  mi.  per  hour.  When  and  where  will  the  fast 
train  be  ordered  to  pass  the  first  one  ? 

6.  Kansas  City  and  Chicago  are  498  mi.  apart.  Where  do  two 
trains  pass,  one  starting  from  Chicago  at  40  mi.  per  hour  at  the  same 
time  another  leaves  Kansas  City  at  50  mi.  j>er  hour? 

7.  A  man  rows  downstream  at  the  rate  of  6  mi.  per  hour  and  rows 
back  at  the  rate  of  3  mi.  per  hour.  How  far  can  he  go  and  return  in 
9  hours? 

[Hint.  Construct  a  line  representing  the  downward  journey  at 
6  mi.  per  hour.  Then,  beginning  at  the  point  which  represents  9  hr., 
construct  the  points  which  represent  his  position  at  each  preceding 
hour,  i.e.  coimt  back  1  hr.  and  up  3  mi.  The  intersection  of  these 
lines  is  the  result  required.] 

8.  A  man  goes  from  Chicago  to  Milwaukee  on  a  train  making  42^ 
mi.  per  hour  and  returns  at  once  on  a  steamer  making  17  mi.  per  hour. 
Find  the  distance  between  the  cities  if  the  trip  requires  7  hr. 

9.  A  freight  train  leaves  a  station  at  12  m.  At  1 :  04  p.m.  a  fast 
train  making  40  mi.  per  hour  leaves  the  same  station  in  the  same  direc- 
tion and  is  ordered  to  pass  the  freight  at  2 :  40  p.m.  What  is  the  rate 
of  the  freight  train,  and  how  far  is  the  point  of  passing  from  the  station  ? 

10.  A  man  rides  a  wheel  into  the  country  at  the  rate  of  8  mi.  per 
hour.  After  riding  a  while  an  accident  occurs  and  he  walks  back  at 
the  rate  of  3  mi.  per  hour,  reaching  home  11  hr.  after  the  start.  How 
far  did  he  go? 


CHAPTER  VII 
COMPUTATION  BY  LOGARITHMS 

101.  Introduction.  Computation  is  ordinarily  carried  on 
by  the  processes  of  multiplication,  division,  raising  to  powers, 
and  extraction  of  roots.  These  arithmetic  processes  are 
often  laborious  by  the  ordinary  methods.  They  may  be 
greatly  abridged  by  the  use  of  tables  of  logarithms  of  num- 
bers. Logarithms  of  the  trigonometric  ratios  are  useful  in 
trigonometric  problems.  The  following  table  illustrates 
the  principles  involved. 

102.  Table  of  Exponents.  The  following  table  of  expo- 
nents gives  the  first  eight  successive  powers  of  10.  The 
table  enables  us  to  reduce  multiplication  and  division  of  these 
numbers  to  addition  and  subtraction,  respectively,  due  to 
the  first  law  of  exponents  : 

Table  of  Exponents  of  10 
100=1.                         103=1000.  10^=1,000,000. 

101  =  10.  10^  =  10,000.  107  =  10,000,000. 

102=  100.  105=  100,000.  108=  100,000,000. 

103.  Extension  of  the  Table  of  Exponents.  It  occurs 
at  once  that  a  more  complete  table  of  exponents  would  be 
quite  useful  in  performing  multiplications  and  divisions. 
The  table  of  §  102  may  be  extended  to  contain  numbers 
less  than  1  by  simply  dividing  1  by  10  repeatedly  and  sub- 

124 


VII,  §  105]  LOGARITHMS  125 

trading  1  from  the  exponent  each  time.  This  conforms 
precisely  to  the  way  this  table  of  exponents  was  made.  Thus 
we  find 

102=100,  101=10,  100=1,  10-1  =^,  10-2  =  3^,  etc. 
The  gaps  in  such  a  table  may  be  filled  in  by  taking  arith- 
metic means  of  successive  exponents  and  geometric  means, 
§  144,  of  the  corresponding  numbers.  For  example,  the 
arithmetic  mean  between  2  and  3  is  2.5.  The  number 
corresponding  to  this  in  the  second  column  of  the  table  of 
exponents  must  then  be  V(100)(1000),  i.e. 
Vl00,000=  100  Vl6  =  316.2, 

which  means  that  (lO)^'^  =316.2  In  practice,  however, 
these  tables  are  computed  by  a  more  rapid  process,  which 
is  too  advanced  to  be  discussed  here. 

104.  Definition  of  a  Logarithm.  The  exponents  of  10 
in  the  first  column  of  the  table  of  §  102  are  called  the  log- 
arithms of  the  numbers  in  the  second  column,  and  10  is 
called  the  base  of  the  system. 

In  general,  the  logarithm  of  a  number  is  the  exponent  by 
which  the  base  is  affected  to  produce  the  given  number.     Thus, 
if  n  is  the  number,  x  the  exponent,  and  b  the  base,  then 
b^=n     and     x  =  logi,  n 

are  equivalent  expressions,  where  logj,  n  is  read,  the  logarithm 
of  n  to  the  base  b. 

For  example,  3^  =  9  is  the  equivalent  of  logs  9  =  2;  4^  =  64 
is  the  equivalent  of  log4  64  =  3.  If  2=^  =  32,  then  x  =  log2  32, 
i.e.  x  =  5. 

105.  Logarithms  to  the  Base  10.  Any  positive  number 
except  1  might  be  used  as  a  base  for  a  system  of  logarithms. 
The  system  with  10  as  base  is  most  generally  used;  it  is 
called  the  common  system  of  logarithms. 


126  MATHEMATICS  [VII,  §  105 

In  this  book,  when  no  base  is  given,  10  will  be  understood 
to  be  the  base ;  and  log  N  will  mean  logio  -^• 

In  general,  logarithms  are  not  integers  but  are  composed 
of  an  integer  and  a  fraction.  The  fractional  part  is  usually 
written  as  a  decimal  fraction.     Thus,  since 

102=100,      i.e.     log  100  =2, 
and  103  =  1000,     i.e.     log  1000  =  3, 

the  logarithm  of  any  number  between  100  and  1000  is  2 
plus  a  certain  decimal.  What  is  wanted  is  a  means  of  find- 
ing this  decimal  part  in  each  instance.  These  decimal  parts 
have  been  determined  and  tabulated.  Such  a  tabulation  is 
called  a  table  of  logarithms.     (See  §  108.) 

106.  Characteristic.  Mantissa.  The  integral  part  of  a 
logarithm  is  called  the  characteristic,  and  the  decimal  part 
is  known  as  the  mantissa. 

Mantissas  are  found  from  a  table,  characteristics  are  de- 
termined by  inspection.  Considerations  of  the  tables  of 
§§  102-103  lead  to  the  following  rule. 

Rule  for  Characteristic.     The  characteristic  of  loga- 
rithms of  numbers  greater  than  unity  is  one  less  than  the  num- 
ber of  digits  to  the  left  of  the  decimal  point. 
For  example, 

log  324  =  2.5105,  log  32.4=1.5105,  log  3.24  =  0.5105 
For  positive  numbers  less  than  unity,  the  characteristic  of 
the  logarithm  is  negative  but  numerically  one  greater  than  the 
number  of  ciphers  immediately  following  the  decimal  point. 
For  example, 

log  .00324=  -3  and  +.5105,  written  7.5105-10,  since 
7-10=-3, 
log  .0324=  -2  and  +.5105,  written  8.5105-10. 
log  .324=  -1  and  +.5105,  written  9.5105-10. 


VII,  §  107]  LOGARITHMS  127 

Observe  that  while  some  characteristics  are  positive  and 
some  are  negative,  the  mantissas  are  always  positive. 

Dividing  a  number  by  10  moves  the  decimal  point  one 
place  to  the  left,  and  subtracts  unity  from  the  logarithm 
of  the  number.  This  does  not  affect  the  mantissa,  but  the 
characteristic  is  decreased  by  unity. 

Similarly,  multiplying  a  number  by  10  moves  the  decimal 
point  one  place  to  the  right,  and  adds  unity  to  the  logarithm 
of  the  number.  The  mantissa  remains  unchanged,  but 
the  characteristic  is  increased  by  unity.  For  example,  if 
(from  a  table)  log  376  =  2.5752, 

then  log  37.6  =  1.5752,  log  3.76  =0.5752, 

log  .376  =  9.5752-10,  log  .0376  =  8.5752-10,  etc. 

107.  Properties  of  Logarithms.  Since  a  logarithm  is,  by 
definition,  an  exponent,  the  rules  for  exponents  apply  to 
logarithms.     Thus,  since 

10'^  •  10*=  10^, 
10-^10*=  io«-», 
(10^^)^=10°^ 
(10«)i/p=10«/P, 

we  have  the  following  rules. 

I.  The  logarithm  of  the  product  of  two  numbers  is  equal  to 
the  sum  of  the  logarithms  of  the  factors  of  this  product,  i.e. 

log  MN=log  M-\-log  N. 
For  example,  log  6  =  log  2+log  3  since  6  =  2  times  3. 

II.  The  logarithm  of  the  quotient  of  two  numbers  is  equal 
to  the  logarithm  of  the  numerator  minus  the  logarithm  of  the 
denominator,  i.e. 

log  —  =  log  M—  log  N. 

N 

For  example,   log  (4/7)  =  log  4 -log  7. 


128  MATHEMATICS  [VII,  §  107 

III.  The  logarithm  of  the  pth  power  of  a  number  is  equal 
to  p  times  the  logarithm  of  the  number,  i.e. 

log  {NP)=plog  N. 
For  example,  log  (2^)  =3  log  2. 

IV.  The  logarithm  of  the  rth  root  of  a  number  is  found  by 
dividing  the  logarithm  of  the  number  by  r,  i.e. 

log  ViV=  -  log  N. 
r 
For  example,  log  (7V2)  =  i  log  7. 

EXERCISES 
Given  log  2  =  .3010,  log  3  =  .4771,  log  5  =  .6990,   find  the  value  of 
each  of  the  following  expressions. 

1.  log  4.  5.   log  10.  9,   log  Vs. 

2.  log  6.  6.   log  12.  10-   log  8/9^ 

4.   log  9.  8.   log  Vs.  '      ^  V34 .  6  y 

108.  The  Table  of  Logarithms.  It  is  supposed  here  that 
a  table  of  mantissas  to  four  decimal  places  is  used.  This 
meets  the  requirements  of  most  of  the  calculating  to  be 
done  in  agriculture,  and  much  of  the  calculating  to  be  done 
in  engineering. 

From  a  table  of  logarithms  we  can  find : 

(a)  the  logarithm  of  a  given  number, 

(b)  the  number  corresponding  to  a  given  logarithm. 

The  general  principles  governing  the  use  of  tables  may 
be  explained  by  the  following  examples.  The  tables  are 
found  in  the  Appendix  of  this  book,  Table  VIII. 

Example  1.     Find  the  logarithm  of  374. 

The  characteristic  is  2.  Now  look  in  the  table  in  the  left-hand 
column,  headed  A^  at  the  top,  for  37.  Then  follow  horizontally  across 
to  the  mantissa  in  the  column  headed  4  at  the  top,  where  5729  is  found. 
This  is  the  required  mantissa.     Thus  log  374  =  2.5729 


VII,  §  108]  LOGARITHMS  129 

Example  2.     Find  log  37. 

For  log  37,  we  look  for  37  in  the  column  headed  N.  Opposite  this, 
in  the  column  headed  0  is  the  mantissa  .5682     Thus  log  37  =  1.5682 

Example  3.     Find  log  32.73 

The  characteristic  is  1.  This  number  cannot  be  found  in  our  table, 
but  we  can  obtain  the  mantissa  for  its  logarithm  by  a  process  called 
interpolation.  This  assumes  that  to  a  small  change  in  the  number  there 
corresponds  a  proportional  change  in  the  mantissa.  The  number  3273 
lies  3/10  of  the  way  from  3270  to  3280.  Hence  the  mantissa  for  log  32.73 
is  similarly  placed  between  log  3270  and  log  3280.  To  obtain  log  32.73, 
we  find  from  the  table 

Mantissa  of  log  3270  =  .5145 

Mantissa  of  log  3280  =  .5159 

Difference  =      14 

Oiu*  desired  mantissa  is  therefore  .5145+3/10  of  14  =  .5149  Hence 
we  have  log  32.73  =  1.5149 

This  may  be  arranged  schematically  as  follows : 

Number  Mantissa 


I  3270  5145  \ 

\  3273  il 


Difference  =  10  \  A|3273  iA  / 14  =  difference, 

3280  5159 

from  which  x:  14  =  3:  10,  i.e.  x  =  3/10  of  14  =  4. 

Example  4.     Fmd  N  if  log  iV  =  0.8485 

This  mantissa  cannot  be  found  in  our  table,  but  we  find  the  mantissas 
next  below  and  above  it,  viz.  8482  and  8488.  These  mantissas  corre- 
spond to  7050  and  7060.  Thus  since  8485  is  3/6  of  the  way  from  8482 
to  8488  we  conclude  that  its  corresponding  number  is  3/6  of  the  way  from 
7050  to  7060,  i.e.  7055.  Since  the  characteristic  is  O  the  number  N 
is  7.055 

Schematically  this  may  be  arranged  as  follows : 


Number 

Mantissa 

1     1  7050 

8482  \     \ 

A 

3] 

Difference  =  10 1       \ 

8485 1/-      /6  =  difference^ 

\l        7060 

8488       l/ 

m  which  x:  10  =  3:6,  i.e.  x  = 

=  5. 

This  added  to  7050  gives  N. 

130  MATHEMATICS  [VII,  §  108 

EXERCISES 

1.  Find  from  the  table  the  logarithms  of  each  of  the  following 
numbers. 

(a)  47.3  (c)    (.063)(417)  (e)  (.0349)1/3 

(6)   .0473  (d)  (.0349)1/2  (/)   (.00349)3/4 

[Hint.  To  find  log  (.0349)1/2,  i.e.  ^  log  (.0349),  take  ^  (8.5428  - 10), 
i.e.  ^  (18.5428-20).  This  is  equal  to  9.2714  To  divide  by  3,  write 
it  28.5428-30.] 

2.  Find  the  logarithms  of  each  of  the  following  numbers. 
(a)  243.6 

[Hint.     Find  first  log  243.0      The  tabular  difference  or  mantissa 

difference  is  18.     The  correction  for  the  fourth  figure  6  is  then  ^jj  of 

18  =  11.     This  added  to  log  243.0  is  the  desired  logarithm.     Notice 

also  that  11  is  found  in  the  6th  column  at  the  right  in  the  same  row  as  24.] 

(6)  3.874  (c)  .8358  {d)  68.72  (e)  4657. 

3.  Find  the  numbers  corresponding  to  each  of  the  following  log- 
arithms. 

(a)  1.4335  Ans.     27.13  (d)  3.4775 

(6)   1.4341  Ans.     27.17  (e)   9.5687-10 

(c)  8.4341-10  Ans.     .02717  (J)   8.7741 

109.  Computation  by  Logarithms.  The  principal  use  of 
logarithms  is  in  shortening  computation.  The  outline  of 
the  work  in  the  following  examples  suggests  similar  arrange- 
ments for  other  exercises.  It  is  desirable  to  make  first  the 
outline  of  the  work  to  be  done,  then  look  up  all  logarithms 
and  enter  them  in  their  proper  places.  After  this  has  been 
done,  the  necessary  additions,  subtractions,  etc.  are  per- 
formed. This  procedure  saves  time,  and  reduces  very  much 
the  chances  of  error. 

Example  1.     Find  25.66  X  18.43 

(1)  log  25.66  =  1.4092 

(2)  log  18.43  =  1.2655 
Sum  of  (1)  and  (2)  =2.6747 

=  log  472.8 
Hence,  the  product  =472.8 


II,  §  109] 

Example  2.    Find 

(1) 

(2) 

(3)  Add  (1)  and  (2) 

(4) 

(5) 

(6)  Add  (4)  and  (5) 

LOGARITHMS  131 

devalue  of  124^X^3843 

.7687  X  49.56 
log  124.8   =  2.0962 
log  .03843=  8.5847-10 

=  10.6809-10 
log  .7687  =  9.8858-10 
log  49.56   =   1.6951 

=  11.5809-10  =  1.5809 
(7)  Subtract  (6)  from  (3)  =9.1000-10 

=  log  (.1259) 
Hence,  the  value  =.1259 

Example  3.     Find  the  value  of  x=(^748)i/3(29.36)^ 

log  x  =  i  log  .3748+2  log  29.36-.i  log  431.5 

(1)  log  .3748  =  9.5738 - 10  =  29.5738  -30  (in  order  to  divide  by  3) 

(2)  i  log  .3748=  9.8579-10 

(3)  log  29.36=   1.4677 

(4)  2  log  29.36=  2.9354 

(5)  Add  (2)  and  (4)  =12.7933-10  =  2.7933 

(6)  log  431.5=   2.6350 

(7)  flog  431.5=   1.3175 

(8)  Subtract  (7)  from  (5)       =11.4758-10 

=  1.4758 
=  log  29.91 
Hence,  x  =  29.91 

MISCELLANEOUS   EXERCISES 
Find  the  value  of  each  of  the  following  expressions  by  means  of 
logarithms. 

J     .  .    (2348)  (.4537)3  (c)    ^§2:83 

■    ^  ^         (83.46) 


/^ri-  "^  (d)  ^/"3547- 

(6)   V13.84  A/TT^i^ 


(1.045)' 

2,  (^)    V3979^-6854.  ^^^   f  4^^^ 

^Q^LS  ^73.96; 

3.  The  amount  of  a  principal  at  compound  interest  for  a  certain  time 
is  given  by  the  formula  ^  ^ p, ^ _^^w 

in  which  A  =the  amount,  P  =  the  principal,  r  =  the  rate,  n  =  the  number 
of  years  or  periods. 


132  MATHEMATICS  [VII,  §  109 

Find  by  use  of  this  formula,  the  amount  of  $598.40  for  15  years  at 
5%  interest  compounded  annually.  Compounded  semiannually. 
Compounded  quarterly. 

[Hint.  For  compounding  semiannually,  use  n  =  twice  the  number 
of  years  and  r  =  half  the  rate.  For  compounding  quarterly,  use  n  =  four 
times  the  number  of  years  and  r  =  one  fourth  the  rate.] 

4.  A  field  whose  successive  vertices  are  A,  B,  C,  D,  E  is  measured 
as  follows:  AB  =  21  rods,  BC  =  S9  rods,  CD  =  23  rods,  DE  =  30  rods, 
EA  =  18  rods,  BE  =  35  rods,  BD  =  38  rods. 

Draw  the  figure  to  scale,  divide  it  into  triangles,  and  compute  the 
area  of  each  triangle  in  acres  from  the  formula  for  the  area  in  acres  of 

a  triangle,  

A_  Vs{s—a){s  —  b){s—c) 
160 
where  a,  h,  c  are,  respectively,  the  sides  of  the  triangle,  and  s  =  \{a  -\-h-\-c) . 

5.  What  is  the  capacity  in  barrels  (40  gallons)  of  a  round  cistern 
12  feet  deep  and  5  feet  in  diameter? 

6.  The  time  of  one  vibration  (one  complete  swing)  of  a  pendulum 
is  given  by  the  formula 

T=WT/i, 

where  I  =  length  of  the  pendulum, 

g  =  attraction  of  gravitation, 
=  32.16  ft.  =980.2  centimeters. 

Find  the  time  of  vibration  of  a  pendulum  21.56"  long.  Of  one 
62.8  centimeters  long. 

Find  the  length  of  a  seconds  pendulum,  i.e.  one  which  makes  one 
vibration  in  one  second. 

7.  How  long  is  a  pendulum  that  vibrates  3  times  in  1  second? 
How  long  is  it  if  it  vibrates  4  times  in  3  seconds?  What  is  the  time 
of  vibration  of  a  pendulum  40  ft.  long? 

8.  A  clock  having  a  seconds  pendulum  gains  2  minutes  per  day. 
How  much  too  short  is  the  pendulum  ? 

9.  The  pendulum  of  a  certain  clock  is  of  brass  and  is  36"  long  at 
70°  F.  A  brass  rod  will  expand  or  contract  approximately  .00001052 
of  its  length  for  each  Fahrenheit  degree  change  in  temperature.  What 
is  the  length  of  this  pendulum  at  40°  F.  ?  How  much  will  the  clock  gain 
in  1  week  due  to  this  change  of  temperature  ? 


VII,  §  110]  LOGARITHMS  133 

10.  A  certain  concrete  sidewalk  578  feet  long  is  laid  in  a  region 
where  the  temperature  varies  from  —20°  F.  to  105°  F,  What  allow- 
ance for  this  change  of  temperature  must  be  made  in  the  length  of  the 
walk  if  concrete  changes  .00000668  of  its  length  for  each  Fahrenheit 
degree  change  of  temperature,  if  the  concrete  is  laid  when  the  tempera- 
ture is  50°  F.? 

11.  Due  to  change  in  temperature  a  substance  made  of  steel  will 
change  .0000065  of  its  length  for  each  change  of  1°  F.  What  space 
should  be  left  between  the  ends  of  steel  rails  60  feet  long  when  laid  in 
a  region  where  the  temperature  varies  from  —4°  F.  to  104°  F.,  if  the 
rails  are  laid  when  the  temperature  is  20°  F.  ? 

110.   The  Slide  Rule.     Introduction.     The  slide  rule  is 

an  apparatus  for  determining  mechanically  products, 
quotients,  powers,  roots.  It  consists  of  two  principal  pieces 
so  made  that  one  may  slide  upon  the  other.  It  is  simply  a 
table  of  logarithms  arranged  so  that  they  may  be  added  or 
subtracted  conveniently,  and  even  more  rapidly  than  with 
the  ordinary  printed  table.  Of  course,  the  logarithms  are 
not  printed  on  the  slide  rule,  but  each  number  simply  stands 
in  the  position  indicated  by  its  logarithm.     This  gives  a 

n \ \ \ \ — I    INI'' 

1  2  S  4  5         6       7      8     9    10 

Fig.  91. — A  Logarithmic  Scale 

scale  which  is  not  uniform  and  which  is  called  a  logarithmic 
scale.  Such  a  scale  is  shown  in  Fig.  91.  This  scale  was 
graduated  as  follows : 

The  distance  PQ  was  divided  into  1000  equal  parts. 
Since  log  2  =  .301,  2  was  placed  at  the  301st  division,  and 
as  log  3  =  .477,  3  was  placed  at  the  477th  division,  etc.  for 
the  succeeding  integers. 

To  multiply  numbers  by  the  use  of  logarithms,  we  add  the 
logarithms  of  the  several  factors  and  find  the  number  cor- 
responding to  the  sum  of  the  logarithms.  Hence,  to  mul- 
tiply two  numbers  by  use  of  the  slide  rule,  it  is  only  necessary 


134 


MATHEMATICS 


[VII,  §  110 


to  place  the  logarithm  of  one  of  the  numbers  on  one  piece  of 
the  rule,  end  to  end  with  the  logarithm  of  the  other  number 
on  the  other  piece  of  the  rule,  and  then  read  off  as  the  product 
the  graduation  mark  at  the  end  of  these  combined  lengths. 

For  example,  on  the  scale  of  Fig,  91  above,  log  2,  which  is 
represented  by  the  distance  from  1  to  2,  added  to  log  3,  the 
distance  from  1  to  3,  gives  log  6,  the  distance  from  1  to  6. 
Thus,  2X3  =  6  on  the  scale. 

Show  in  the  same  way  that  2X5  =  10.  Also  that  (2.5)  X 
(3.5)  =  ?  Determine  it  first  on  the  scale  and  compare  with 
the  result  obtained  by  actual  multiplication.  Similarly 
(2.5)X(3.2)=  ? 

111.  Description  and  History  of  the  Slide  Rule.  One  part 
of  the  rule  is  called  the  slide.  It  slides  along  a  groove  in 
the  center  of  the  other  part  called  the  stock.  Some  rules 
carry  a  runner  for  convenience  in  continued  multiplication 
and  division  and  squares  and  square  roots.  This  is  merely 
a  sliding  frame  with  a  wire  or  line  on  it  perpendicular  to  the 
length  of  the  rule. 

The  device  of  having  one  piece  slide  on  the  other  was 
invented  by  William  Ought  red  between  1620  and  1630, 
soon  after  the  invention  of  logarithms  by  Napier  in  1614. 


Fig.  92.  —  The  Mannheim  Arrangement 

Practically  all  slide  rules  in  present  use  are  Mannheim 
rules.  They  are  so  named,  not  from  any  manufacturer,  but 
from  Lieutenant  Mannheim  of  the  French  Army,  who 
devised  the  present  arrangement  of  the  scales  about  1850. 
Some  higher  priced  rules  have  modifications  of  the  Mannheim 


VII,  §  113]  LOGARITHMS  135 

arrangement,  but  the  fundamental  principles  involved  are 
the  same  as  in  the  regular  Mannheim  rule.  Fig.  92  shows 
the  regular  Mannheim  arrangement.* 

112.  The  Scales.  Two  scales  are  engraved  along  the 
upper  edge  of  the  groove.  One,  labeled  A,  is  on  the  stock, 
the  other,  labeled  B,  is  on  the  slide.  These  scales  are 
identical.  The  slide  simply  makes  it  possible  to  add  graphi- 
cally the  segments  on  the  scales. 

Along  the  lower  edge  of  the  groove  are  also  two  scales. 
One,  C,  is  on  the  slide,  the  other,  D,  is  on  the  stock.  These 
two  scales  are  likewise  identical. 

Scales  A  and  B  range  from  1  to  100  while  C  and  D  range 
only  from  1  to  10.  The  range  on  one  set  being  smaller,  the 
number  of  divisions  can  be  increased.  Thus  C  and  D  give 
greater  accuracy  in  reading  them  since  there  are  more 
divisions.  Notice  also  that  every  number  on  the  A  and 
B  scales  is  the  square  of  every  number  below  it  on  the  C 
and  D  scales,  due  to  the  fact  that  the  unit  on  one  is  twice 
as  great  as  on  the  other.  This  gives  us  at  once  a  table  of 
squares  and  square  roots.  The  runner  enables  one  to  find 
corresponding  numbers  on  the  upper  and  lower  scales,  and 
also  to  hold  a  given  position  during  a  repeated  operation. 

113.  Operations  with  the  Slide  Rule.  To  use  a  slide 
rule  successfully,  one  must  acquire  the  virtues  of  speed  and 
accuracy.  This  requires  practice,  since  the  accuracy  of  a 
result  often  requires  the  estimation  of  a  number  falling  be- 
tween the  lines  of  division. 

*  Fig,  92  is  reproduced  on  a  larger  scale  on  the  first  fly  leaf  at  the  back 
of  the  book.  By  cutting  out  this  leaf  and  carefully  cutting  up  the  figure, 
a  slide  rule  can  be  made  by  the  student.  It  will  not  be  very  accurate,  but 
it  will  suffice  to  illustrate  the  principles. 

Slide  rules,  varying  from  the  inexpensive  pasteboard  kind  to  the  higher 
priced  wood  and  celluloid  ones  used  by  engineers,  may  be  had  at  practically 
all  stores  which  carry  draftsmen's  supplies. 


136  MATHEMATICS  [VII,  §  113 

A  skillful  operator,  with  a  10-inch  slide  rule,  can  always 
secure  results  accurate  to  three  significant  figures.  This  is 
accurate  enough  for  most  of  the  purposes  of  applied  science. 

The  beginner  should  use  small  numbers  until  he  becomes 
familiar  with  the  operations. 

114.  Multiplication.  Multiply  2  by  3.  Move  the  shde 
so  as  to  put  1  of  the  B  scale  on  2  of  the  A  scale ;  then  above 
3  on  the  B  scale  read  the  product  6  on  the  A  scale.  This  is 
simply  adding  logarithms. 

Notice  that  the  1  at  either  end  of  the  B  scale  may  he  used. 
Try  both  ways.  Use  that  1  which  brings  the  second  factor 
under  scale  A . 

For  example,  multiply  65  by  3  using  CD  scales.  If  we 
set  the  left  end  1  of  C  on  65  D,  then  3  C  is  off  the  scale  D. 
In  this  case  put  the  right  end  1  of  C  on  65  D  which  brings 
3  C  under  195. 

The  decimal  point  is  placed  by  inspection,  i.e.  by  making 
an  approximate  mental  computation  to  determine  the  num- 
ber of  integers  or  the  number  of  decimal  places. 

For  example,  multiply  18.5  by  2.8.  Set  1  C  on  185  D 
and  under  28  C  read  518  D.  But  the  product  is  roughly 
18X3  =  54 ;  therefore  we  know  that  there  are  only  two  places 
before  the  decimal  point  in  the  product  of  18.5  and  2.8 
Thus  518  read  on  D  means  51.8 

115.  Division.  Divide  6  by  2.  Set  2  C  on  6  D,  under 
1  C  read  the  quotient  3  on  D.  This  means  merely  sub- 
tracting logarithms. 

Also,  divide  7.2  by  2.5.  Set  25  C  on  72  D,  under  1  C  read 
the  quotient  2.9  on  D.     Therefore, 

To  divide  one  number  by  another,  set  the  divisor  on  scale  C 
on  the  dividend  on  scale  D,  and  under  1  C  read  the  quotient 
on  D. 


VII,  §  116]  LOGARITHMS  137 

116.   Combined  Multiplications  and  Divisions.     Find  the 

^^^^^  ^^  (25.2)  (9.6) 

8.4 

Set  84  C  on  252  D,  and  under  9.6  C  read  the  quotient  28.8 
on  D. 

Notice  that  84  (7  on  252  D  gives  the  quotient  under  1  C 
but  as  this  is  to  be  multipHed  by  96,  1  C  is  already  on  this 
quotient  and  hence  we  have  only  to  read  the  product  on 
scale  D  under  9.6  C. 

An  important  application  of  this  is  finding  the  fourth  ter)n 
of  a  proportion.     For  example,  from  the  proportion 

8.4  :  25.2  =  9.6  :x, 
we  find 

^_  (25.2)  (9.6) 
"^  SA 

The  rule  then  to  find  the  fourth  term  is  obvious,  viz. : 

To  find  the  fourth  term  of  a  proportion,  set  the  first  term  on 
the  second,  and  under  the  third  term  read  the  fourth. 

In  continued  multiplications  and  divisions,  the  runner 
with  its  perpendicular  indicator  is  quite  convenient. 

Example  1.     Find  the  value  of  2  X  3  X  4. 

Set  1  C  on  2  D,  set  runner  on  3  C  (marking  this  product).     Now 

set  1  C  at  right  end  on  runner  and  under  4  C  read  24  on  D.     Thus 

2X3X4  =  24. 

56 
Example  2.     Find  the  value  of  - — -• 

7X4 

Set  7  C  on  56  D,  set  runner  on  1  C  (marking  the  quotient),  set  4  C  on 

the  runner  and  under  1  C  read  the  result  2  on  D. 

Example  3.     Find  the  value  of  ?|^- 

17X12 

Set  17  C  on  24  D,  set  runner  on  51  C,  set  12  C  on  rimner,  under  1  C 

read  result  6  on  D. 


138  MATHEMATICS  [VII,  §  117 

117.  Squares  and  Square  Roots.  Scales  C  and  D  were 
constructed  on  a  unit  twice  as  large  as  that  for  scales  A  and 
B.  Accordingly,  every  number  on  the  A  and  B  scales  is 
the  square  of  the  number  vertically  below  it  on  the  C  and 
D  scales.  Thus  above  2  C  we  find  4  A,  above  3  is  9,  above 
25  is  625,  etc.     Thus, 

To  square  any  number,  find  the  number  on  scale  C  and 
read  its  square  vertically  above  it  on  scale  B. 

To  extract  the  square  root  of  a  number,  find  the  number  on 
scale  B  and  read  its  square  root  vertically  below  it  on  scale  C. 

Example  1.     Find  the  value  of  16  V3. 

Set  1  C  on  3  ^,  under  16  C  read  the  result  27.7  on  D. 

Example  2.     Find  the  value  of  I6/V5. 

i^  =  16V5. 
V5      5 
Set  5  C  on  5  A,  under  16  C  read  result  7.16  on  D. 

Example  3.  Find  the  area  of  a  circle  whose  radius  is  3  ft.  Set 
1  C  on  3  D,  above  tt  on  B  read  the  area  28.3  sq.  ft.,  on  A. 

EXERCISES 

Find  the  value  of  each  of  the  following  expressions. 


1. 
2. 

3. 

(4.67)  (8.32). 

(.341)(3.86). 

31.5               . 
16.2 

,     (97)  (15) 
23 

5. 

7. 
8. 

849           g     (18.6)  (3.4) 
31.6                    3.1416 

27V19 
37 

(47.2)(18.1)(.47). 

on 

9.   Compute  W, 
white  oak  beams, 

the  safe  load  in  tons, 
from  the  formula 

when  uniformly  distributed 

W  = 

2hd^ 
"3   I  ' 

' 

if  b  (the  breadth  in  inches)  is  3,  d  (the  depth  in  inches)  is  8,  and  I  (the 
distance  in  inches  between  supports)  is  192. 

10.   Find  the  area  of  a  circle  whose  radius  is  75  miUimeters. 


CHAPTER  VIII 
THE  PROGRESSIONS 

118.  Introduction.  Progressions  are  among  the  oldest 
topics  of  all  mathematics.  Egyptian  scholars  had  studied 
them  thousands  of  years  before  the  Christian  era.  By 
2000  B.C.  they  had  attained  to  a  considerable  knowledge  of 
them,  as  evidenced  in  the  papyrus  of  Ahmes. 

Progressions  are  important,  not  only  in  the  solution  of 
numerical  problems,  but  also  in  theoretical  developments. 
For  example,  out  of  the  association  of  arithmetic  and  geo- 
metric progressions  grew  the  subject  of  logarithms. 

119.  Arithmetic  Progression.  An  arithmetic  progression 
(written  A.  P.)  is  a  succession  of  terms  so  related  that  any 
one,  after  the  first,  is  obtained  from  the  preceding  one  by 
adding  a  constant  number.  This  constant  is  known  as  the 
common  difference.     Thus 

2,  5,  8,  11,  •••,  with  a  common  difference  3, 
5,  1,  —3,  —7,  •••,  with  a  common  difference  —4, 
a,  a-\-x,  aH-2  x,  •••,  with  a  common  difference  x, 
are  arithmetic  progressions. 

Is  2,  3,  5,  4,  7,  11,    ••,  an  arithmetic  progression  ? 

120.  Terms  of  an  Arithmetic  Progression.  In  such  a 
succession  of  terms  as  just  described,  we  are  interested  in 

the  first  term,  known  as  a, 
the  common  difference,  known  as  dj 
the  last  or  nth  term,  known  as  ly 
the  number  of  terms,  known  as  n, 
and  the  sum  of  the  terms,  known  as  s. 

139 


140  MATHEMATICS  [VITI,  §  120 

Therefore,  we  may  write  the  arithmetic  progression  in 
general  as  follows. 
1st  term       2d  term       3d  term       17th  term      nth  or  last  term 

a,  a-\-d,       a-\-2d,    ■',  a-\-lQ  d,  ■■■,     a-{-(n—l)d. 

It  thus  appears  that  any  term  is  equal  to  the  first  term,  plus 

the  common  difference  multiplied  by  the  number  of  the  term 

less  one,  i.e. 

l=a-\-{n-l)d. 

For  example,  the  17th  term  =  a+(17-l)d  =  a+16  d. 

EXERCISES 

1.  Given  the  A.  P.  3,  7,  11,  •••.     What  is  the  common  difference? 
What  is  the  10th  term? 

Begin  with  the  10th  or  last  term  and  write  the  series  backwards. 

2.  Do  as  in  Ex.  1  for  the  A.  P.  1,  4,  7,   ••  to  10  terms. 

3.  Do  as  in  Ex,  1  for  the  A.  P.  a,  a-\-x,  a-\-2  x,  •••  to  10  terms. 

4.  If  I  is  the  last  term  and  d  the  common  difference,  write  the  A.  P. 
backwards. 

[Hint.     If  I  is  the  last  term,  then  l—d  is  the  next  to  the  last,  etc.] 

5.  Write  the  A.  P.  which  exhibits  the  strokes  of  a  clock  striking  the 
hours. 

121.   Siun  of  the  Terms  of  an  Arithmetic  Progression. 

The  simi  of  an  A.  P.,  i.e.  the  sum  of  all  the  terms  taken,  is 
found  as  follows. 

s=^a+{a-\-d)-\-{a-\-2  d)-\-  ••' to  ntermsj 
or,  written  backwards, 

s  =  l^{l-d)  +  {l-2d)-\-  •••  to  n  terms. 
Addition  of  corresponding  terms  gives 

2  s=  (a+Z)  +  (a+0+  "•  taken  n  times  =  n(a+0. 
Thus,        s=^ia+l). 


VIII,  §  121]  THE  PROGRESSIONS  141 

Example  1.  Find  the  total  number  of  strokes  made  by  a  clock  in 
striking  12  hours. 

Here  a  =  l,  1  =  12,  n  =  12;  therefore  s  =  V(l+12)  =78. 

Example  2.  What  is  the  total  price  paid  for  a  lot  bought  on  the 
installment  plan  of  25  f^  the  first  week,  50^  the  second  week,  75  jf  the 
third  week,  etc. ;  to  continue  for  one  year  of  52  weeks? 

Here  a  =  25,  d  =  25,  n  =  52;  hence  the  last  term  is 
Z  =  25 +  (51)25  =  1300  ff, 

8  =  ^  (a +0  =  26(25  + 1300)  =  34,450ff  =  $344.50. 

EXERCISES 

Find  the  last  term  and  the  sum  of  all  of  the  terms  in  each  of  the 
following  A.  P.'s. 

1.  5,  9,  13,  •••  to  20  terms.  Ans.     Z  =  81,  s  =  860. 

2.  1,  3,  5,  7,  •••  to  25  terms. 

3.  2i,  3|,  5,  •••  to  15  terms. 

4.  I,  -J,  -h  '••  to  16  terms. 

5.  Find  the  sum  of  the  first  50  even  integers  beginning  with  0. 

6.  A  row  of  twenty  trees  is  set  at  intervals  of  fifteen  feet,  beginning 
at  a  point  forty  feet  from  a  house.  How  far  is  the  last  one  from  the 
house  ?  If  a  man  carries  a  bucket  of  water  from  a  well  by  the  house  to 
each  tree,  how  far  does  he  walk  ? 

7.  Write  an  A.  P.  of  nine  terms  for  which  the  first  term  is  4  and 
the  last  term  20. 

[Hint.     Use  l  =  a  +  {n  —  l)d  where  a  =  4,  Z  =  20,  etc.] 

8.  How  far  apart  must  9  plants  be  set  to  be  equally  spaced  in  a 
row  16  ft.  long?  Ans.     2  ft. 

9.  If  there  are  sixteen  plants  in  a  row  40  feet  long,  how  far  apart 
are  they? 

10.  If  the  sum  of  an  A.. P.  is  18f,  the  first  term  |,  the  number  of 
terms  10,  find  I  and  d. 

11.  A  man  deposits  $75  in  a  savings  bank.  How  much  must  this 
and  later  deposits  be  increased  each  year  so  that  in  10  years  his  total 
deposit  shall  amount  to  $1875,  not  counting  interest? 


142  MATHEMATICS  [VIII,  §  121 

12.  A  farmer  has  $100  set  aside  for  drilling  a  well.  How  deep  could 
he  have  it  drilled  at  25^  for  the  first  foot  and  for  each  additional  foot 
1  ^  more  than  the  preceding  one  ? 

13.  A  freely  falling  body  descends  16  ft.  the  first  second  and  ap- 
proximately 32  ft.  more  each  second  thereafter  than  in  the  preceding 
second.  How  deep  is  a  well  to  the  bottom  of  which  a  stone  falls  in 
3  sec.  ?     In  6  sec.  ?     How  far  does  it  fall  in  the  sixth  second  ? 

14.  A  bomb  is  released  from  an  airplane  1  mile  high.  How  long 
does  it  take  the  bomb  to  reach  the  ground  ?  How  far  did  it  go  the  last 
second  ? 

If  the  airplane  is  going  75  miles  per  hour,  how  far  horizontally  in 
advance  of  its  starting  point  does  the  bomb  hit  the  ground  ? 

15.  According  to  the  law  of  falling  bodies  as  given  in  Ex.  13,  what 
is  the  relation  between  distance,  s,  and  time,  t,  i.e.  between  the  sum  of 
the  A.  P.  and  the  number  of  terms? 

122.  Geometric  Progression.  A  geometric  progression 
(written  G.  P.)  is  a  succession  of  terms  so  related  that  any 
term,  after  the  first,  is  obtained  by  multiplying  the  preceding 
term  by  a  constant  number  called  the  ratio.     Thus 

2,  8,  32,  •••,  with  ratio  4, 
and  —2,  6,  —18,  54,  •••,  with  ratio  —3, 

are  geometric  progressions. 

What  is  the  ratio  for  each  of  the  following  geometric  progressions : 
9,3,1,-1,...? 
4,  -12,  36,  -108,-? 

123.  Terms  of  a  Geometric  Progression.  In  a  geometric 
progression,  we  are  interested  in 

the  first  term,  known  as  a, 

the  ratio,  known  as  r, 

the  last  or  nth  term,  known  as  I, 

the  number  of  terms,  known  as  n, 

the  sum  of  the  terms,  known  as  s. 


VIII,  §  124]  THE  PROGRESSIONS  143 

Therefore,  we  may  write,  generally,  the  series  of  terms  of 
a  G.  P.  as  follows. 

1st  term        2d  term        3d  term      17th  term      nth  or  last  term 
a,  ar,  ar^,  •••  ar^^,  •••  ar^^. 

The  exponent  of  r  in  any  term  is  one  less  than  the  number 
of  that  term.     The  nth  or  last  term  is 

For  example,  the  17th  term  is  ar^^. 

EXERCISES 

Which  of  the  following  sequences  of  numbers  represent  geometric 
progressions  ?     Find  the  ratio  and  the  last  term  in  each  such  progression. 
!•   ij  i»  1>  •••  to  9  terms. 

2.  2,  3,  5,  4,  7,  •••  to  6  terms. 

3.  2,  i  ,V  •••  to  6  terms. 

4.  7,  .7,  .77,  •••  to  5  terms. 

5.  72,  .72,  .0072,  •••  to  5  terms. 

6.  (a+6),  (a+6)2,  (a+ft)',  •••  to  6  terms. 

7.  (a+fe),  (a2-62),  (^a^-ab^-a^+b^),  —  to  4  terms. 

8.  V2,  2,  Vs,  •••  to  7  terms. 

9.  If  a  and  ar'^  are  the  end  terms,  respectively,  of  a  G.  P.  of  three 
terms,  what  is  the  middle  term?  What  is  the  middle  term  if  4  and  9 
are  the  end  terms  of  a  G.  P.  of  three  terms? 

10.   If  4  is  the  first  term  of  a  G.  P.  and  ^^  is  the  fourth  term,  find  the 
second  and  third  terms. 

[Hint.     Let  o  =  4,  and  ar^  =  y  ;  find  v.] 

124.   Sum  of  the  Terms  of  a  Geometric  Progression.    By 

definition  the  sum  of  n  terms  of  a  G.  P.  is 

s  =  a-\-ar-\-ar^-\-  •••  +ar"-i. 

Therefore       rs  =  ar-]-ar'^-{-ar^-\-  •••  -\-ar^. 

Subtracting  these  gives 

s  —  rs  =  a  —  ar"^  =  a  —  rar^~^  =  a—rlo 

rrtu  a  —  ar"      a  —  rl 

Thus  s  = = 

1  — r        1  — r 


144  MATHEMATICS  [VIII,  §  124 

Example.  In  the  G.  P.  6,  2,  f,  •••  find  the  6th  term  and  the  sum  of 
6  terms. 

Here  a  =  6,  r  =  |,  n  =  6,  l  =  aj^-^  =  6{^)^  =  6^lj  =  j\. 
Hence  s  is  given  either  by  the  formula 

„_a-ar^_6-6a)6_728 
1-r  1-^  81  ' 

or  by  the  formula 

o_a-rl  ^Q-^'  A -728. 
1-r         1-i  81* 

EXERCISES 

1.  If  a  =  3,  r  =  ^,  n  =  8,  find  i  and  s. 

2.  If  a  =  6,  n  =  6,  Z=  — ff,  find  r  and  s. 

3.  How  many  terms  of  the  G.  P. 

1^+2^+4^+  .- 

are  necessary  so  that  the  sum  shall  be  at  least  $40? 

Ans.     12  terms. 

4.  A  blacksmith  agrees  to  shoe  a  horse  as  follows :  ^|if  for  the  first 
nail  driven,  ^^  for  the  next,  1^  for  the  next,  etc.,  for  the  8  nails  of  one 
shoe,  and  a  like  amount  for  each  of  the  other  three  shoes.  What  does 
he  receive  for  the  work?  Ans.     $2.55 

5.  If  the  first  term  of  a  G.  P.  of  5  terms  is  2  and  the  last  term  1250, 
find  the  ratio  and  the  sum  of  5  terms. 

6.  Find  the  sum  of  7  terms  of  1+2-1+2-2+—. 

7.  If  a  =  2,  n  =  5,  Z  =  1250,  find  rand  s. 

8.  If  a  =  17,  Z  =  459,  n  =  4,  find  r  and  s  and  write  the  series. 

9.  The  common  housefly  will  lay,  incubate,  and  mature  a  litter 
of  eggs  about  every  3^  weeks.  What  will  be  the  number  of  descendants 
of  one  female  fly  in  14  weeks  if  there  are  150  to  a  Utter,  evenly  distributed 
as  to  sex? 

10.  Find  the  amount  of  $1  at  4%  compound  interest  for  1  year. 
For  2  years.     For  3  years.     For  4  years. 

[Hint.  The  rate  is  4%  =  .04  and  the  amount  is  $1.04  for  one  year. 
This  is  the  principal  for  the  second  year.  The  amount,  therefore,  for 
the  second  year  is  (1.04)  times  (1.04)  =  (1.04)2.] 

Do  these  results  form  a  geometric  progression? 


VIII,  §  124]  THE  PROGRESSIONS  145 

11.  Find  the  amount  of  p  dollars  at  rate  r  compound  interest  for 
1  year.     For  2  years.     For  3  years.     For  n  years. 

Ans.     The  amount  for  n  years  =  p(l  +r)'*. 

[Hint.  The  amount  of  $1  for  1  year  at  rate  r  is  $1(1+0  and  the 
amount  of  $p  is  therefore  p  times  this,  i.e.  $p(l+r).  The  amount  for 
the  second  year  is  therefore  (1+r)  times  the  new  principal  $p(l+r), 
i.e.  it  is  p(l+r)2.] 

12.  Find  the  amount  of  $50  for  8  years  at  5%  compound  interest. 
[Hint.    The  computation  is  shortened  by  use  of  logarithms.] 

Ans.    $50(1.05)'. 

13.  A  man  deposits  $50  each  year  in  a  building  and  loan  association 
at  5%  compound  interest.  What  is  the  total  amount  of  interest  and 
deposits  at  the  end  of  6  yr.  ? 

[Hint.  The  first  deposit  is  compounded  at  5%  for  6  yr.  and  thus 
amounts  to  50(1.05)**.  The  second  deposit  amounts  to  50(1.05)^ 
being  on  interest  for  5  yr.  The  next  deposit  amounts  to  50(1.05)*, 
etc.     The  total  amount  for  the  six  years  is  thus 

50(1.05) +50(1.05)2-1- ...  +50(1.05)'. 

But  this  is  a  G.  P.  of  six  terms.    Show  that  its  sum  is 

50(1.05)  [il^5|^]. 

Compute  this  amount  by  use  of  logarithms.] 

14.  Find  the  total  amount  of  a  yearly  deposit  of  p  dollars  com- 
pounded at  rate  r  for  n  years.  rn  _i_».^n_n 

Ans.     p(l+r)[ii±^^^ ij. 

15.  On  a  certain  life  insurance  endowment  policy  for  $1000,  the 
insured  agrees  to  pay  $47.68  each  year  for  20  years,  and  he  is  to  receive 
$1000  at  the  end  of  the  period.  What  is  the  total  amount  of  these 
deposits  compounded  annually  at  3^%?  What  does  the  company 
obtain  at  the  end  of  the  period  for  carrying  the  risk,  if  it  earns  just  3|% 
on  money  in  its  possession?  Ans.     Amt.  =$1395.44 

16.  A  man  buys  a  machine  for  $150.  What  amount  of  money 
should  be  deposited  each  year  to  draw  4%  compound  interest  so  that 
the  amount  of  these  deposits  would  purchase  a  similar  new  machine 
at  the  end  of  12  years?  Ans.     $9.60 

[Hint.     Use  the  formula  of  Ex.  14,  solving  for  p.] 


146  MATHEMATICS  [VIII,  §  124 

17.  What  equal  annual  deposit  at  4%  compound  interest  will 
amount  to  $300  in  15  years? 

18.  What  amount  of  money  could  profitably  be  expended  for  a 
harvesting  machine  which  will  last  15  years,  if  it  saves  annual  labor 
bills  of  $40,  money  being  worth  5%. 

[Hint.  The  total  saving,  with  interest  from  the  end  of  each  year, 
is  given  by  the  sum  of  the  geometric  progression 

,     s  =  40(1.05)14+40(1.05)13+..^  +40  =  40[(1.05)l5-l]-^(.05). 
But  this  is  the  saving  at  the  end  oi  15  years.     The  principal  p  invested 
now  would  produce  in  15  years  the  amount  p(1.05)i^.     It  follows  that 
p{1.05y^  =  s.     Solve  for  p,  using  logarithms.]         Ans.     p  =  $415.20 

125.  Infinite  Geometric  Progressions.  Thus  far  we 
have  considered  only  those  progressions  which  have  a  limited 
number  of  terms.  There  are  many  problems  and  phenomena 
whose  discussion  calls  for  a  series  of  an  unlimited  number 
of  terms.  For  example,  suppose  an  elastic  ball  falls  4  feet 
and  rebounds  half  of  that  distance,  then  falls  2  feet  and  re- 
bounds half  of  that  distance,  etc.,  rebounding  each  time  half 
of  the  distance  fallen.  The  entire  distance  fallen  would 
be  the  sum  of  the  series 

4+2+l+i+i+  •••to  infinity. 
This  is  a  G.  P.  in  which  the  ratio  is  less  than  unity  and  the 
number  of  terms  is  unlimited. 

If,  in  a  G.  P.,  r<l,  each  term  is  smaller  than  the  preced- 
ing, the  last  term  I  approaches  zero  as  the  number  of  terms 

increases,  and  the  formula  s= becomes,  when  Z  =  0, 

1  —  r 

o     1-    'J    r        a  —  ro        a 
S  =  limit  of  s= = 

1—r        1—r 
Example.     Find  the  sum  of  the  infinite  series 

l+^+i+i+TV+-- 

Here  a  =  l,  r  =  ^  and  aS= -^  = -^  =2. 

1—r      1—^ 

This  simply  means  that  as  the  number  of  terms  of  this  series  in- 
creases, the  sum  of  them  differs  less  and  less  from  2. 


VIII,  §  125]  THE  PROGRESSIONS  147 

EXERCISES 

1.  Find  the  sum  of  the  following  infinite  geometric  progressions. 

(a)  l+l+i  +  2V+iV+-- 

(c)    1-i+i-i+i^--. 

The  progression  (d)  is  (c)  transformed  by  subtracting  successive 
pairs  of  terms. 

2.  What  common  fraction  gives  rise  to  the  repeating  decimal  ,44  •  •  •  ? 
[Hint.     Find  the  sum  of  the  series 

.4+.04+.004+-..] 

3.  What  common  fraction  gives  rise  to  .777---?  To  .2727  •••? 
To  .315315  •••? 

4.  A  farm  returns  a  net  income  of  $800  per  year.  Considering 
that  money  is  worth  5%  and  that  this  return  can  be  kept  up  indefinitely, 
what  is  the  farm  worth?  How  much  is  it  worth  if  money  is  worth  4% ? 
6%? 

[Hint.  The  amount  of  $1.00  for  1  year  at  5%  is  $1.05.  Therefore 
the  present  value  of  an  amount  due  in  1  year  is  that  amount  divided 
by  1.05.  The  amount  of  $1.00  for  two  years  at  5%  is  $(1.05)2,  and 
the  present  value  of  $1  two  years  hence  is  1/(1.05)2.  xhe  present  value 
of  the  first  return  of  $800  due  in  1  year  is  therefore  $800/(1.05) ;  the 
present  value  of  the  second  return  due  in  two  years  is  $800/ (1.05) 2; 
etc.,  indefinitely.    The  sum  of  the  series  800/(1.05)  +800/(1.05)2+  ...  jg 

800  _.  fi  _    1   \^800    .    .05  ^800       1.05  ^  800  _  800  .  ^qq 
1.05  ■  V       1.05/      1.05  *  1.05      1.05       .05       .05        5 

This  gives  the  following  rule  for  determining  the  value  of  property 
that  yields  a  fixed  income :  Multiply  the  income  by  100  and  divide  by 
the  number  of  per  cent  in  the  assumed  rate  of  interest.] 

5.  Find  the  value  of  a  farm  that  yields  a  net  income  of  $2400,  with 
money  at  5%.  At  8%.  Does  a  higher  interest  rate  give  lower  or  higher 
valuations  ? 

6.  Suppose  the  farm  in  Ex.  4  produces  $800  net  return  only  on 
alternate  years,  i.e.  it  Hes  fallow  every  other  year.  What  is  its  value 
with  money  at  5%? 


148  MATHEMATICS  [VIII,  §  125 

7.  A  ball  falls  6  ft.  and  at  each  fall  rebounds  2/5  of  the  distance. 
Find  the  distance  the  ball  has  traveled  before  coming  to  rest. 

8.  Find  the  time  it  takes  the  ball  in  the  previous  exercise  to  come 
to  rest  using  s  =  l6t^  as  the  law  of  falling  bodies  where  s  =  distance 
fallen  in  t  seconds. 

9.  The  mid-points  of  the  sides  of  a  square  12"  on  a  side  are  joined 
to  form  a  smaller  square.  Then  the  mid-points  of  the  sides  of  this 
smaller  square  are  joined  to  form  a  still  smaller  square.  This  process 
is  continued  indefinitely.  What  is  the  sum  of  the  perimeters  of  these 
squares  ? 


CHAPTER  IX 

COMPOUND  INTEREST— ANNUITIES— 
DEPRECIATION 

126.  Compound  Interest.  When  the  interest  on  a  note 
or  an  account  is  added  to  the  principal  as  it  becomes  due, 
and  the  new  amount  draws  interest,  the  owner  is  said  to 
receive  compound  interest.  Interest  is  computed  in  this 
way  by  building  and  loan  associations,  life  insurance  com- 
panies, large  corporations,  etc.,  since  they  can  reinvest  at 
once  any  deposits,  receipts,  premiums,  etc. 

Compound  interest  is  no  longer  allowed  on  notes.  But 
if  any  one  collects  his  interest  when  due,  and  at  once  reinvests 
it,  he  virtually  has  the  advantage  of  compound  interest. 

In  computing  interest  semiannually,  the  rule  is  to  take 
twice  the  time  and  one  half  the  rate,  since  the  payments  are 
made  twice  as  often,  and  are  only  half  as  large.  For  com- 
puting interest  payable  quarterly,  take  four  times  the  time 
and  one  fourth  the  rate.  Proceed  in  a  similar  manner  for 
any  specified  number  of  payments  per  year. 

127.  Amount  of  $p  at  Rate  r  Compoimded  for  n  Years. 
The  amount  of  $1  at  rate  r  for  1  year  is  $(l+r).  Then  the 
amountof  $(l+r)  for  the  next  year  is  $(l+r)(l+r)  =  $(l+r)2. 
The  amount  of  $(H-r)2  for  the  next  year  is  $(l+r)2(l+0 
=  $(l+r)^.  This  is  the  amount  of  $1  for  3  years.  For 
4  years  it  is  $(l-|-r)*,  etc.  For  n  years  the  amount  is 
$(l+r)".     Then  the  amount  of  $p  for  n  years  at  rate  r  is 

(1)  A=p(l+r)\ 

149 


150  MATHEMATICS  [IX,  §  127 

Example.  Find  the  amount  of  $500  deposited  in  a  savings  bank 
at  5%  annually,  the  interest  compounded  semiannually  for  2  years. 

The  semiannual  rate  is  2|%,  and  is  computed  for  four  periods. 

First  Solution:  As  in  Exs.  10,  11,  §  124,  the  amount  of  $500  for 
4  periods    @  2^%  =  $500(1.025)^.     This  may  be  computed  by  ordinary 
arithmetic ;  but  the  work  is  shortened  by  the  use  of  logarithms, 
log  500  =  2.6990 
4  log  (1.025)=   .0428 

Sum  =  2.7418  =  log  (551.90). 

Thus  the  compound  amount  is  $551.90 

Note.  In  operations  involving  numbers  of  five  or  six  figures,  a  five- 
place  or  a  seven-place  table  of  logarithms  should  be  used.  If  the  time 
interval  is  over  twenty  years,  at  least  a  five-place  table  is  desirable. 

Second  Solution:  By  a  compound  interest  table  (see  Appendix), 
the  amount  of  $1.00  for  4  periods  @  2^%  =  $1.10381  and  the  amount 
of  $500  for  4  periods   @  2§%  =  $500($1. 10381)  =$551.90 

EXERCISES 

1.  Find  the  amount  of  $150  at  5%  interest  compounded  annually 
for  3  years. 

2.  Use  logarithms  to  find  the  amount  of  $50  at  6%  compound 
interest  for  5  years.  Ans.     $66.91 

[Hint.     The  rate  is  6%,  hence  r  =  .06,  and  1  +r  is  1.06] 

3.  Solve  Ex.  2  if  the  interest  is  compounded  semiannually. 
[Hint.     Use  10  periods  at  3%.] 

4.  A  father  deposits  $25  in  a  building  and  loan  association  for  his 
son  at  birth.  It  draws  5%  compound  interest  payable  semiannually. 
What  is  the  amount  due  the  son  when  he  is  21  years  old? 

5.  When  a  son  is  10  years  old,  his  father  agrees  to  deposit  for  him 
in  a  building  and  loan  association,  beginning  at  once  and  on  each  suc- 
ceeding birthday,  a  sum  of  money  equal  to  his  age.  The  deposit  is 
to  bear  4%  compound  interest  payable  semiannually.  What  is  the 
amount  to  his  credit  when  he  is  21  years  old?  Ans.     $226.32 

[Hint.  The  first  deposit  of  $10  bears  interest  for  11  years,  etc. 
The  son  is  also  to  be  credited  with  $21  on  his  twenty-first  birthday.] 

6.  A  man  deposits  $50  at  the  beginning  of  each  year  in  a  building 
and  loan  association  which  pays  5%  per  year,  payable  semiannually. 
If  the  interest  and  the  several  principals  are  allowed  to  accumulate, 
what  will  be  the  total  amount  at  the  end  of  10  years? 


IX,  §  129]  COMPOUND  INTEREST  151 

128.  Annuities.  An  annuity  is  a  series  of  equal  payments 
made  at  equal  intervals  of  time  during  a  fixed  term  of  years. 

The  first  payment  is  supposed  to  be  made  at  the  end  of 
the  first  year  unless  otherwise  specified. 

Annuities  are  usually  computed  at  compound  interest 
and  are  therefore  applications  of  geometric  series  (§  124). 
The  whole  subject  of  annuities  is  of  great  fundamental 
importance,  since  its  practical  applications  include  farm 
loans,  rents,  leases,  pensions,  dowers,  life-estates,  salaries, 
life  insurance,  old  age  insurance,  disability  insurance,  and 
many  other  similar  financial  operations. 

129.  Farm  Loans.  Under  the  recent  Farm  Loan  Act 
instituted  by  the  Federal  Government,  a  farmer  may  borrow 
from  the  Government  on  long  time  and  at  a  nominal  rate 
of  interest.  A  problem  which  then  arises  with  him  is  to 
pay  off  this  loan,  both  principal  and  interest,  by  a  series 
of  equal  payments  extending  over  a  period  of  years  specified 
in  the  agreement. 

For  example,  a  farmer  buys  a  farm  for  $10,000  and  pays 
for  it  by  a  loan  from  the  Government.  He  may  wish  to 
know  now  how  large  a  payment  to  make  each  year,  say  for 
30  years,  so  that  at  the  end  of  that  time  the  loan  will  be 
cancelled.  The  problem  may  be  solved  by  considering 
how  large  an  annuity  $10,000  will  purchase  at  5%  interest, 
for  example,  for  the  30  years.  The  answer  to  this  particular 
problem  is  $650.50  per  year. 

Notice  also  that  this  $650.50  per  year  pays  the  interest 
on  the  loan  each  year,  $500,  and  besides  pays  $150.50  on 
the  principal.  This  payment  of  $150.50  each  year  at  5% 
for  30  years  must  therefore  accumulate  at  compound  interest 
to  precisely  $10,000  at  the  end  of  the  period.  This  it  does 
and  the  loan  is  accordingly  cancelled. 


152  MATHEMATICS  [IX,  §  130 

130.  Problem  A.  Amount  of  an  Annuity.  To  find  the 
amount  of  $1  payable  at  the  end  of  each  year  for  n  years. 

The  first  payment  made  at  the  end  of  the  first  year  draws 
interest  at  rate  r,  for  n— 1  years,  and  amounts,  by  §  127, 
to  (l+r)»-i. 

The  second  payment,  similarly,  for  n  — 2  years  amounts 
to  (l+r)»-2. 


The  next  to  last  payment,  for  1  year,  amounts  to  (1+r)^ 
The  last  payment  at  the  end  of  the  term  is  1. 
The  sum  of  these  amounts  is  the  amount  desired.     It  is 
usually  called  s^,.     Thus 

s^=l  +  (H-r)  +  (l+r)2+  ...  (H-r)--i. 

By  §  124,  the  sum  of  this  series  of  terms  is 

(2)        .,=  (1+^):^. 

r 

If  the  payment  each  time  is  p,  the  total  amount  is 

(3)  Ps.=  p(i^±^} 

If  payments  are  made  at  the  beginning  of  each  year,  as  is 
dften  the  case,  the  amount  at  the  end  of  n  years  is  denoted 
hy^s. 

The  amount  is  the  same  as  if  the  preceding  amount  s^ 
were  allowed  to  run  another  year ;  hence 

(4)  ^„5=(l+rK-i=(l+r)(^^+^^)"-^)- 

Formulas  (2),  (3),  and  (4)  can  be  computed  from  a  com- 
pound interest  table  (see  Appendix),  since  such  a  table  gives 
the  value  of  (l+r)«,  the  amount  of  1  compounded  annually 
for  n  years  at  rate  r. 


IX,  §  130]  COMPOUND  INTEREST  153 

Example  1.  Find  the  accumulated  value  of  an  annuity  of  $500 
a  year  at  5%  compounded  annually  for  15  years,  payments  being  made 
at  the  end  of  each  year. 

First  Solution  :   From  equation  (3),  the  desired  amount  is 


,^^=$500[(l±f^]  =$500[(1Q5V--1]. 


From  a  four-place  table  of  logarithms, 

15  log  (1.05)  =  .3180  =log  (2.0795), 
that  is  (1.05)^5  =  2.0795 

Thus,  ^^=$500^'^"^^^"^  =  $10,795.* 

"•  .05 

Second  Solution  :   By  a  compound  interest  table,  the  value  of 
(1.05)15  =  2.07893, 
and  s-  =  $10,789.30 

Example  2.     Calculate  the  accumulation  in  the  previous  example 

if  the  payments  were  made  at  the  beginning  of  each  year. 

From  equation  (3),  --s,  the  amount  desired  in  this  case  =  (1 +r)s^. 

Therefore, 

^  =  (1.05)  ($10,789.30)  =$11,328.76 

EXERCISES 

1.  Find  the  accumulation  of  $150.50  per  year  at  5%  for  30  years, 
payments  being  made  at  the  end  of  the  year.  Ans.     $9999.05 

2.  A  man  deposits  at  the  beginning  of  each  year  $150.50  in  a  build- 
ing and  loan  association  at  5%  compound  interest.  What  is  the  total 
value  at  the  end  of  30  years?  Ans.     $10,499. 

3.  What  will  a  farmer  save  in  30  years,  if  his  yearly  saving  is  $300 
and  is  compounded  at  5%?  Ans.     $19,931.65 

4.  Find  the  amount  of  an  annuity  of  $100  at  the  end  of  each  year 
for  10  yr.  at  5%  interest,  compounded  semiannually. 

Ans.     $1261.47 

5.  Find  the  amount  of  $50  at  the  end  of  every  six  months  for  10 
years  at  5%  interest,  compounded  semiannually.        Ans.     $1277.23 


*  By  use  of  five-place  table  of  logarithms,  «^  in  this  example  is  equal  to 
$10,789.30 


154  MATHEMATICS  [IX,  §  131 

131.  Problem  B.    Present  Worth  of  an  Annuity.     To 

find  the  present  value  of  an  annuity  of  1  per  annum  for  n  years. 
The  present  value,  x,  of  1  due  in  n  years  is  the  sum  which 
will  amount  to  1  in  n  years  with  compound  interest  at  rate  r. 
Therefore 

a;(H-r)"=l,    i.e.    x  = 


(l+r)» 
Thus 
the  present  value  of  the  first  payment,  receivable  in  1  year, 

isl/(l+r); 

the  present  value  of  the   second   payment,   receivable   in 
2  years,  is  l/(l+r)2; 

the  present   value  of  the  nth  payment,   receivable  in  n 
years,  is  1/(1 +r)"- 

But  the  present  value  of  the  whole  annuity  is  the  sum  of 
the  present  values  of  the  several  payments.     This  sum  may 
be  denoted  by  a^.    Thus 
,.,      ^   _     1  1  1 

This  is  a  geometric  series  whose  first  term  is  1/(1 -fr)  and 
whose  ratio  is  l/(l+r).  Hence,  by  §  124,  we  find  after 
simplifying, 

(6)         _l-(l  +  r)-_(l+r)»-l  1       _      5^     . 

^^^     ^"^  r  r  (l  +  r)»      (1  +  r)" 

If  each  payment  is  p  instead  of  1,  the  present  value  of 
such  an  annuity  is  merely  p  times  a^  in  equation  (6). 

Equation  (6),  in  its  final  form,  shows  that  the  present 
value  of  an  annuity  of  SI  is  merely  the  amount  of  an  annuity 
for  SI  for  n  years  divided  by  the  amount  of  SI  at  compound 
interest  for  n  years.  In  this  form  it  may  be  computed  either 
by  arithmetic,  or,  far  more  easily,  from  a  compound  interest 
table. 


IX,  §  132]  COMPOUND  INTEREST  155 

132.  Problem  C.  Present  Price  of  an  Annuity.  To  find 
that  annual  payment  for  n  years  which  $1  will  purchase.  Let 
p  =  the  annual  payment  continued  for  n  years,  at  rate  r. 
Its  present  or  cash  value  is  its  cost,  i.e.  $1  in  this  case. 
Therefore,  by  equation  (6),  §  131, 

g>i_J(l+r)--l  1 

Hence  the  annual  payment,  p,  for  n  years,  which  $1  will 
purchase  is  ,.  .  (\^rY 

^"(l+r)"-l' 
and  the  annual  payment,  P,  which  %A  will  purchase  is 

^^  (l+r)— 1 

Example  1.     What  annual  payment  for  10  years  will  $500  purchase, 

if  money  is  worth  5% ? 

SoTTTTioN-    p-  500  •  (.05) (1.05)^"  _  (500) (.05) (1.62890) 
SOLUTION,    r-       (io5)io_i  (1.62890) -1      ' 

from  a  compound  interest  table.     Therefore  P  =  $64.75 

EXERCISES 

1.  What  is  the  present  value  of  $600  due  in  8  years  if  money  is 
worth  4%  compounded  annually?  Ans.     $438.41 

2.  Solve  the  previous  example  if  the  interest  is  computed  at  4% 
compounded  semiannually.  Ans.     $437.07 

3.  Find  the  present  value  of  an  annuity  of  $53  receivable  at  the  end 
of  each  year  for  20  years,  at  5%,  (o)  by  logarithms  and  (6),  §  131, 
(6)  by  a  compound  interest  table.  Ans.     $660.50 

4.  A  man  buys  a  farm  and  agrees  to  pay  for  it  $1000  in  cash  and 
$500  at  the  end  of  each  year  for  10  years.  If  money  is  worth  6%,  what 
would  be  the  equivalent  cash  price  of  the  farm  ?         Ans.     $4680.44 

5.  What  annual  payments  continued  for  10  years  are  equal  to 
$1000  cash,  if  money  is  worth  5%?  Ans.     $129.50 

[Hint.     Let    p  =  the    annual    payment.     Then    from     (6),    since 

»-=«-■        «ooo=p[a^g-.^]. 

from  which  p  may  be  computed.] 


156  MATHEMATICS  [IX,  §  132 

6.  What  annual  payments  for  20  years  can  be  purchased  for  $500, 
money  being  valued  at  5%  ?  Ans.     $40.12 

7.  What  semiannual  payments  continued  for  20  years  can  be 
purchasedfor  $500,  at  5%?  Ans.     $19.92 

8.  A  man  buys  a  farm  for  $10,000  and  wishes  to  pay  for  it  by  a 
series  of  equal  annual  payments  of  such  amount  as  to  pay  both  principal 
and  interest  in  30  years  at  5%.  Find  the  amount  of  the  annual  pay- 
ment. Ans.     $650.50 

[Hint.     Find  what  annual  payment  $10,000  will  purchase.] 

9.  A  man  secures  $2400  for  the  purchase  of  farm  equipment  by  a 
loan  from  the  government.  He  agrees  to  pay  $20  per  month,  including 
6%  interest  on  all  sums  remaining  unpaid.  How  long  will  it  take  to 
pay  off  the  loan  and  what  will  be  the  amount  of  the  last  payment  ? 

Ans.     184  months ;  last  payment,  $14.26 

10.  A  farmer  wishes  to  provide  an  income  for  old  age.  He  esti- 
mates that  at  age  30  years  he  will  have  35  years  of  productive  activity 
ahead  of  him,  and  that  he  can  save  $250  per  year  during  that  time. 
This  accumulation  at  5%  compound  interest  at  age  65,  together  per- 
haps with  $2500  cash  available  from  the  sale  of  his  farm  and  tools, 
will  purchase  what  annual  payments  for  15  years,  if  money  is  worth  5%  ? 

Ans.     $2416.27 

11.  Solve  the  previous  exercise,  estimating  that  the  farmer  can  siave 
$100  per  year  for  30  years  at  5%  and  that  he  will  use  this  accumulation 
at  age  60  years,  plus  $2500,  to  purchase  an  income  for  20  years  at  5%. 
What  annual  payments  can  he  thus  purchase?  Ans.     $733.73 

133.  Problem  D.  Sinking  Fund.  To  find  the  annual 
payment  that  will  amount  to  $A  in  n  years.  Such  annual 
payments  made  to  accumulate  a  given  amount  at  a  given 
time  constitute  a  sinking  fund. 

Let  p  denote  the  annual  payment,  and  A  the  amount  at 
the  end  of  n  years.     Then,  by  (3),  §  130,  the  amount  is 

(8)  pi^±^  =  A, 

r 

whence 


IX,  §  134]  COMPOUND  INTEREST  157 

EXERCISES 

1.  A  farmer  wishes  to  provide  $1500  for  the  replacement  of  ma- 
chinery ten  years  hence.  How  much  must  he  set  aside  each  year  at 
6%  compound  interest  to  meet  this  contingency?        Ans.     $119.26 

2.  How  much  would  the  farmer  of  Ex.  1  need  to  set  aside  every  six 
months  to  provide  for  the  same  amount,  if  the  interest  is  compounded 
semiannually?  Ans.     $58.72 

3.  A  township  issues  bonds  for  $100,000,  payable  at  the  end  of  25 
years,  for  road  improvements.  What  sum  must  be  set  aside  each  six 
months  to  meet  the  payment  of  these  bonds  when  due,  if  the  sums  set 
aside  bear  4%  interest?  Ans.     $1182.30 

4.  A  man  owes  a  note  of  $640  due  at  the  end  of  5  years  with  simple 
interest  at  6%.  He  proposes  to  meet  the  payment  of  both  principal 
and  interest  at  the  end  of  the  time  by  depositing  a  certain  amount 
every  six  months  in  a  savings  association,  at  5%  compound  interest. 
Compute  the  amount  to  be  thus  deposited.  Ans.     $74.26 

134.  Perpetuities.  Capitalization.  When  the  payments 
of  an  annuity  are  continued  indefinitely,  it  is  called  a  per- 
petuity. Perpetuities  are  important  in  practical  business 
affairs.  An  annual  income  from  the  rent  of  property,  from 
an  endowment  fund,  or  from  a  fixed  salary,  has  a  definite 
cash  equivalent.  Likewise,  an  annual  outgo  continued 
indefinitely  has  a  cash  equivalent.  The  process  of  finding 
such  cash  equivalents  is  called  capitalization. 

Problem  E.  To  find  the  present  value  of  a  perpetuity  whose 
payments  are  made  annually. 

Let  p  =  the  annual  payment  due  1  yr.  hence,  and  thereafter 
annually  indefinitely. 

The  present  value  of  the  first  payment  due  in  1  year  is 

p/a+r); 

The  present  value  of  the  second  payment  due  in  2  years 

isp/(l+r)2; 

The  present  value  of  the  third  payment  due  in  3  years 

isp/(l+r)3; 


158  MATHEMATICS  [IX,  §  134 

and  so  forth  indefinitely.  The  present  value  of  the  per- 
petuity is  the  sum  of  these  present  values  of  the  separate 
payments.     If  s^  denote  this  sum,  we  have 

This  is  an  infinite  geometric  series  whose  first  term  is 
p/(l+r),  and  whose  ratio  is  1/(1 +r).  Hence,  by  §  125  its 
sum  is  J. 


(11) 


1  +  r     _p 
1         r 


l+r 

EXERCISES 

1.  If  money  is  worth  5%,  what  is  the  cash  equivalent  of  a  yearly 
income  of  $1500?  Ans.     $30,000. 

2.  What  is  the  value  of  a  farm  which  returns  an  income  of  $750 
per  year,  continued  indefinitely,  money  being  worth  6%? 

Ans.     $12,500. 

3.  What  is  the  capitalized  value  of  a  fixed  salary  of  $3000  a  year, 
with  money  at  6% ?  Ans.     $50,000. 

4.  A  railroad  company  pays  a  flagman  $1200  per  year.  What  sum 
could  the  company  afford  to  pay  to  install  automatic  machinery  to  do 
away  with  the  services  of  the  flagman,  with  money  at  4%? 

Ans.     $30,000. 

5.  Is  farm  land  which  rents  for  $9.00  per  acre  selling  too  high  at 
$250  per  acre?  What  would  be  a  reasonable  rate  of  interest  to  allow 
in  this  case  ? 

135.  Depreciation.  Depreciation  is  the  decrease  in  value 
of  any  perishable  article  which  results  from  use  and  from 
advancing  age.  It  may  result  also  from  progress  in  the 
arts  and  in  methods  of  manufacture  which  renders  an  article 
obsolete,  or  out  of  date.  Articles  may  become  inadequate, 
and  may  have  to  be  replaced,  in  the  course  of  time,  by  others 
better  adapted  to  the  requirements.    Thus,  broadly  speak- 


IX,  §  137]  COMPOUND  INTEREST  159 

ing,  depreciation  includes  wear  and  tear,  obsolescence,  and 
inadequacy. 

Annual  depreciation  is  the  annual  theoretical  decrease  in 
value,  expressed  in  money. 

136.  Methods  of  Determining  Depreciation  Charges. 
Various  methods  have  been  advocated  to  determine  how 
much  shall  be  charged  to  depreciation  for  a  given  article, 
or  plant,  or  machine.  It  is  convenient  to  measure  deprecia- 
tion in  terms  of  time,  notwithstanding  the  fact  that  it  does 
not  necessarily  depend  wholly  upon  the  passage  of  time. 
Periods  of  time  are  merely  a  convenient  way  of  comparing 
the  rapidity  of  depreciation  with  the  rapidity  of  other 
occurrences.  In  §§  137-139  we  shall  consider  three  methods 
of  computation. 

137.  The  Straight  Line  Method.  This  method  proceeds 
upon  the  theory  that  the  annual  depreciation  of  an  article 
is  uniform  throughout  its  probable  period  of  life.  For 
example,  if  a  harvester  costs  $120,  and  has  a  probable  life 
of  10  years  with  no  salvage  value  at  the  end  of  that  time,  the 
annual  depreciation  is  merely  one  tenth  of  $120.  If  the 
salvage  value  is  $15,  then  the  annual  charge  for  depreciation 
would  be  ($120- $15)/10=  $10.50. 

Let  X  =  the  annual  charge  for  depreciation  on  an  article, 
y=its  value  or  cost, 

y  =  its  salvage  value  after  a  period  of  n  years. 
Then,  by  the  straight  line  method,  we  have 

(12)  x=i:^. 

n 

EXERCISES 

1.  Show  that  all  considerations  of  interest  are  disregarded  in  the 
straight  line  formula  for  estimating  the  depreciation  charge. 


160 


MATHEMATICS 


[IX,  §  137 


2.  Draw  a  graph  to  show  the  value  of  the  harvester  of  §  137  at  various 
times  after  its  purchase.     Why  is  (12)  called  the  straight  Une  formula? 

3.  A  house  is  worth  $8000  and  its  estimated  life  is  40  years.  Allow- 
ing $400  salvage  value,  what  should  be  allowed  each  year  for  deprecia- 
tion, using  the  formula  (12)  ?     Draw  a  graph. 

4.  What  is  the  life  of  a  wagon  which  depreciates  about  5%  of  its 
original  value  each  year  ?     Draw  a  graph.  Ans.     20  years. 

5.  A  Minnesota  farm  bulletin  shows  the  following  average  depre- 
ciation of  machinery,  in  per  cents  of  the  original  cost. 


Machine 

%  Depreciation 

Machine 

%  Depreciation 

Threshing  Outfit  . 

12.0 

Gang  Plows     .     . 

7.40 

Hay  Loaders    .     . 

11.78 

Gasoline  Engines 

7.35 

Manure  Spreaders 

11.67 

Corn  Cultivators 

7.23 

Corn  Binders  .     . 

10.03 

Heavy  Horses 

6.17 

Harrows      .     .     . 

8.72 

Wagons  .... 

4.89 

Grain  Binders  .     . 

7.91 

Fanning  Mill  .     . 

4.58 

Determine  the  life  of  each  machine  by  the  straight  line  formula, 
estimating  the  salvage  value  at  5%  of  the  cost  in  each  case. 

Estimate  the  value  at  the  end  of  6  years  of  a  corn  harvester  costing 
originally  $125. 

6.  What  is  the  annual  charge  for  depreciation  and  what  is  the 
probable  life  of  a  hay  loader  that  costs  $110,  and  that  may  be  sold  for 
$30  after  7  years'  usage  ? 

7.  If  the  cost  of  a  machine  is  $100,  its  probable  life  20  years,  and  its 
salvage  value  zero,  what  is  the  annual  charge  for  depreciation,  using 
formula  (12)  ? 

Construct  a  graph  for  this  information,  using  time  in  years  on  the 
horizontal  axis,  and  depreciation  in  dollars  on  the  vertical  axis. 

Construct  a  second  graph,  using  time  as  before,  but  actual  value  of 
the  machine  on  the  vertical  axis. 

138.  The  Reducing  Balance  Method.  This  plan  deducts 
for  depreciation,  at  the  end  of  each  year,  a  certain  constant 
percentage  of  the  value  of  the  article  at  the  beginning  of 
that  year.  This  value  itself  decreases  each  year.  Thus  if  an 
article  costs  $100  and  if  the  depreciation  each  year  is  10% 


IX,  §  139]  COMPOUND  INTEREST  161 

of  the  value  at  the  beginning  of  that  year,  its  value  at  the 
end  of  the  fourth  year  is  found  as  follows  : 

10%of  $100=$10; 
hence  first  balance  =  $100 -$10  =$90; 

10%  of  $90  =$9; 
hence       second  balance  =  $90  —  $9  =  $81 ; 

10%  of  $81  =  $8. 10; 
hence  third  balance  =  $81 -$8.10  =$72.90; 

10%  of  $72.90  =  $7.29; 
hence         fourth  balance  =  $72.90 -$7.29  =  $64.61 

This  process  may  be  continued  till  the  balance  reaches  a 
given  salvage  value.  It  will  be  noticed  that  by  this  method 
the  charge  for  depreciation  is  high  in  the  first  years  of  life  of 
an  article  and  comparatively  low  in  the  later  years  of  its  life. 

139.  The  Reducing  Balance  Formula.  Let  x  denote  the 
constant  fractional  rate  of  depreciation  which  is  to  be  de- 
ducted each  year  from  the  balance ;  F,  the  original  invest- 
ment ;  y ,  the  salvage  value ;  and  n,  the  number  of  years 
before  the  salvage  value  is  reached.  Then,  deducting  x 
of  V  from  V\  we  have 

the  value  at  the  end  of  1  yr.  =  F  —  Fa;  =  F(l  —  a:) ; 
thevalueattheendof2yr.=  F(l-a:)-F(l-x)a:  =  F(l-x)2; 
the  value  at  the  end  of  3  yr.  =  F(l  —  xY ; 

the  value  at  the  end  of  n  yr.  =  F(l  —  x)". 

But  this  nth  balance  is  the  salvage  value,  F'.     Thus 

(13)  V{l-xY=r,     i.e. 

(14)  x=l-S/f. 
and 

(15)  ,^lo,V'-logV, 

log{l-x) 


162  MATHEMATICS  [IX,  §  139 

Example  1.     What  is  the  value  at  the  end  of  the  fifth  year  of  an 
automobile  costing  $500,  if  depreciation  each  year  is  15%  of  the  value 
at  the  beginning  of  the  year? 
By  (13),         V'  =  Va-x)- 

=  $500(1 -.15)5 

=  $500(.4436)  =$221.80 

Example  2.  In  what  time  will  an  automobile  costing  $500  de- 
preciate to  $100,  allowing  15%  reduction  each  year  on  the  value  at  the 
beginning  of  the  year? 

By  (15),  „=logl00-log500 

log  (1-.15) 

^  2.0000-2.6990  _  -(.6990) 
9.9294-10  -(.0706) 

=  9.9  years  =  10  years,  approximately. 

140.  The  Sinking  Fund  Method.  By  the  sinking  fund 
method,  a  fixed  sum  of  money  is  set  aside  each  year,  and  is 
allo'wed  to  accumulate  at  compound  interest.  It  derives 
its  name  from  the  fact  that  it  employs  the  principles  ordi- 
narily used  in  estabHshing  sinking  funds  to  Hquidate  in- 
debtedness and  for  purposes  of  replacements. 

Let  us  assume  that  a  machine  costs  $1000,  that  its  probable 
life  is  25  years,  and  that  its  salvage  value  is  zero.  The 
problem  then  is  to  find  what  sum  of  money  must  be  set 
aside  each  year  in  order  that  at  the  end  of  the  25  years  the 
fund  will  have  accumulated  to  $1000  at  5%  compound 
interest.  The  annual  fixed  sum  thus  set  aside  in  any  year 
plus  the  interest  accumulated  in  that  year,  represents  the 
depreciation  charge  for  that  year.  This  problem  has  al- 
ready been  discussed  and  solved  in  §  133.  In  the  preceding 
example  the  annual  payment  x  is  to  be  found  that  will 
accumulate  to  $1000  in  25  years  at  5%  compound  interest. 
Sy  (9)>  §  133,  the  annual  payment  is 

Vr 
^     {l-\-r)--l' 


IX,  §  140]  COMPOUND  INTEREST  163 

Hence,  for  the  example  just  cited, 

(1000)  (.05)  ^  120.952  =  $20.95  per  year. 
(1.05)25-1  ^     ^ 

EXERCISES 

1.  A  horse  3  years  old  costs  $250  and  at  age  28  has  depreciated 
to  $10  in  value.  Find  the  annual  per  cent  of  depreciation  by  the  re- 
ducing balance  formula. 

2.  Find  the  value  of  the  horse  of  Ex.  1,  by  the  reducing  balance 
method,  at  age  5,  at  age  8,  at  age  13,  at  age  18,  at  age  25,  using  the  per 
cent  of  depreciation  found  in  Ex.  1.  Draw  a  graph  to  represent  these 
values. 

Does  this  computation  of  depreciation  agree  with  the  facts  for  the 
usual  horse,  i.e.  is  depreciation  of  a  horse  uniform  over  a  period  of  years? 
Is  it  large  at  any  time  ?     When  is  it  largest  ? 

3.  Use  the  straight  line  method  to  discuss  Exs.  1  and  2. 

4.  What  is  the  Ufe  of  a  grain  harvester  costing  $240,  allowing  $10 
salvage  value,  and  annual  depreciation  of  10%,  using  the  reducing 
balance  method?  ^ns.     30  years. 

5.  Find  by  the  reducing  balance  method  the  value  of  the  harvester 
of  Ex.  4  at  the  end  of  2  years,  5  years,  10  years,  15  years,  20  years, 
25  years.     Plot  the  graph. 

Discuss  these  values,  and  decide  where  the  depreciation  is  largest, 
where  smallest,  where  it  should  be  largest,  etc. 

When  is  this  binder  worth  $125,  including  salvage? 

6.  Use  the  straight  line  method  to  solve  and  discuss  Exs.  4  and  5. 

7.  A  farmer  buys  a  manure  spreader  for  $120.  After  using  it  for 
3  years,  he  sells  it  for  $75.  Discuss  the  annual  depreciation  charge 
(a)  by  the  straight  Une  method,  (6)  by  the  reducing  balance  method. 

8.  What  annual  payment  should  be  set  aside  by  the  sinking  fund 
method  to  replace  a  kitchen  stove  in  15  years,  if  it  is  valued  at  $75, 
and  if  money  is  worth  5%? 

9.  What  amount  should  be  set  aside  annually  to  replace  an  auto- 
mobile that  costs  $1500,  at  the  end  of  five  years,  if  money  is  worth  5%? 

10.  A  farmer  purchases  an  automobile  for  $2000,  a  harvester  for 
$500,  and  a  thresher  for  $1000,  He  assumes  that  the  automobile 
will  last  for  6  years,  the  harvester  10  years,  and  the  thresher  12  years. 
How  much  should  be  set  aside  annually  to  repurchase  these  machines 
when  worn  out,  not  counting  salvage,  if  money  is  worth  5%? 


164  MATHEMATICS  [IX,  §  141 

141.   Summary  of  Interest  Functions  and  Other  Formulas. 

In  the  following  formulas,  n  denotes  the  number  of  years 
or  periods  of  time,  and  r  denotes  the  rate  of  interest. 

I.    i4  =  />(!+ r)",    the    accumulation    of    %p    compounded 
n  years. 

II-   P^  f.  ,    x«>  ^^6  present  worth  of  $A  due  in  n  years. 

(1+r)" 

(l+r)"— 1 
III.    5^=  ^        ^ ,  the  accumulation  of  $1  paid  at  end 

of  each  year  for  n  years. 

^        ^ ,  accumulation  of  $1  paid  at 

the  beginning  of  each  year  for  n  years. 

For  the  accumulation  of  $p  paid  annually,  multiply  III 
and  IV  by  p. 

V.  a^=  ^        ^ • — ,   the   present   worth   of   an 

annual  payment  of  $1  continued  for  n  years.     For  an  annual 
payment  of  $p,  multiply  by  p. 

VI.  J^=    (l+r)"r       ^^     annual  income  which  $1  will 
fl^      (l+r)»-l' 

purchase. 

1  r 

VII.  — =- ,   the   annual    payment   which   will 

Sn\     (l+r)»-l 

accumulate  to  $1. 

VIII.  5^=-,  the  present  value  of  an  annual  payment 
of  %p  continued  indefinitely. 

IX.  x= ,  straight  line  depredation  from   V  to   V 

n 

in  n  years. 


IX,  §  141]  COMPOUND  INTEREST  165 

X.  a;  =  1  —  ^J— ,  reducing  balance  depreciation  from  V  to 

F'  in  n  years. 

Vr 

XI.  x= ; ,  sinking  fund  allowance  for  replacement 

(1+r)"— 1 

of  the  original  value,  V,  in  n  years. 

MISCELLANEOUS   EXERCISES 

1.  Find  the  amount  of  $1000  at  4%  compound  interest  for  10  years. 

2.  Compute  the  amount  of  $1000  at  4%  compound  interest  for 
10  years,  compounded  semiannually. 

3.  Compute  the  amount  of  $1000  at  4%  compound  interest  for 
10  years  and  6  months. 

[Hint.  (1.04)ioi  may  be  computed  readily  by  logarithms,  but  the 
commercial  practice  is  to  compute  the  compound  interest  to  the  end  of 
the  last  even  year,  and  then  find  the  simple  interest  on  this  amount  for 
the  remaining  fraction  of  a  year.] 

4.  What  sum  at  compound  interest  for  6  years  at  4%  will  accumu- 
late to  $1000? 

State  this  exercise  in  terms  of  present  worth. 

5.  What  rate  at  compound  interest  is  a  man  receiving  when  $1000 
will  amount  to  $1935.28  in  15  years?  • 

6.  In  what  time  will  a  sum  of  money  at  5%  compound  interest 
double  itself  ?  Ans.     14.2  years. 

7.  Find  the  amount  of  $1  at  5%  compound  interest  for  1  year, 
2  years,  3  years,  4  years,  etc.,  to  10  years.  Construct  a  graph  showing 
this  information, 

8.  Construct  a  graph  showing  the  amount  of  $1  for  10  years  as 
the  rate  varies.     Use  2%,  2|%,  3%,  etc.,  up  to  6%. 

9.  How  much  will  a  man  have  saved  at  the  end  of  10  years  if  he 
deposits  $60  semiannually  in  a  savings  bank,  and  interest  is  com- 
pounded at  5%  semiannually? 

10.  What  would  be  the  cash  value  of  a  farm  for  which  a  man  agreed 
to  pay  $1000  a  year  for  10  years,  if  money  is  worth  6%  ?  How  much  if 
money  is  worth  4|%?  Ans.    $7360.09;  $7912.72 


166  MATHEMATICS  [IX,  §  141 

11.  A  tractor  that  costs  $1500  must  be  replaced  in  10  years'  time. 
If  money  can  be  accumulated  at  5%  interest,  what  sum  should  be  set 
aside  each  year  to  meet  the  cost  of  a  new  one  at  the  end  of  the  time  ? 

Ans.     $119.26 

12.  What  annual  income  for  30  years  will  $10,000  purchase,  if 
money  is  worth  4|%?  Ans.     $613.92 

13.  What  annual  income  for  15  years  will  $10,000  purchase,  if 
money  is  worth  4^%?     What,  if  money  is  worth  4%? 

Ans.     $931.14;  $899.41 

14.  If  money  is  worth  5%,  compare  the  following  offers  for  a  farm  : 

(a)  A  rental  of  $600  per  year  for  20  years,  or  a  cash  price  of  $10,000. 

(b)  A  rental  of  $600  per  year  for  10  years  plus  $700  a  year  for  the 
following  10  years,  or  a  cash  price  of  $12,500. 


CHAPTER  X 

AVERAGES  AND  MIXTURES 

142.  Mixtures.  In  preparing  many  compounds,  foods, 
fertilizers,  concrete,  etc.,  many  different  mixtures  of  sub- 
stances or  of  two  varieties  of  the  same  substance,  are  in 
common  use. 

The  simplest  case  is  that  of  mixing  two  varieties  of  the 
same  substance  that  differ  in  percentage  of  strength  of  some 
important  ingredient.  As  an  illustration,  we  shall  first 
show  how  to  determine  the  proper  amounts  of  different 
grades  of  milk  to  form  milk  of  a  desired  standard  of  quality. 

143.  Dairy  Problems.  Milk  is  standardized  according 
to  its  butter  fat  content.  Such  a  standard  milk  may  be 
obtained  by  adding  milk  of  a  higher  per  cent  of  butter  fat 
when  the  per  cent  is  too  low  or  by  the  addition  of  skim  milk 
when  the  per  cent  is  too  high.  Cream  is  also  used  in  the 
standardizing  process.  Water  cannot  be  used,  since  it 
contains  none  of  the  other  elements  necessary  in  milk,  such 
as  sugar,  casein,  etc.  The  amount  of  water  in  milk  is  quickly 
determined  by  the  specific  gravity  test.* 

Ordinary  legal  milk  is  about  a  3%  milk,  i.e.  3%  of  its 
weight  is  butter  fat.  Skim  milk  is  theoretically  a  0%  milk, 
but  it  usually  contains  a  small  trace  of  butter  fat,  from 
.1%  to  .5%. 


*  See  gallon  in  the  Table  of  Equivalents,  Table  V,  Appendix,  where  the 
weight  of  milk  is  shown. 

167 


168  MATHEMATICS  \K,  §  143 

Example.  It  is  desired  to  add  skim  milk  to  800  lb.  of  milk  con- 
taining 4.5%  of  butter  fat  to  form  a  standardized  milk  containing  4% 
of  butter  fat.     How  many  pounds  of  skim  milk  must  be  added? 

Solution.     Let  x  be  the  number  of  pounds  of  milk  after  adding 
skim  milk.     Then  the  amount  of  butter  fat  is  .04  x ;  hence 
.04  X  =  (.045)  (800)  lb.  =  36  lb., 

^^  a:=^=900  1b. 

.04 

It  follows  that  100  lb.  of  skim  milk  should  be  added. 
Still  another  method  is  given  in  §  148. 

EXERCISES 

1.  How  many  pounds  of  skim  milk  must  be  added  to  200  pounds 
of  6%  milk  to  standardize  it  to  4%  milk? 

2.  How  many  pounds  of  skim  milk  must  be  added  to  500  pounds 
of  5.8%  milk  to  standardize  it  to  3%  milk? 

3.  How  much  skim  milk  must  be  removed  from  600  pounds  of 
3.2%  milk  to  standardize  it  to  4%  milk? 

4.  Given  200  lb.  of  4%  milk.  Show  how  to  standardize  it  to  5% 
milk  by  removing  skim  milk. 

5.  How  much  20%  cream  can  you  make  from  50  lb.  of  6%  milk? 
How  much  is  it  worth  if  butter  sells  at  56  ff  per  pound  —  one  pound  of 
butter  fat  being  sufficient  for  116f  %  of  its  weight  in  butter? 

6.  In  standardizing  milk  containing  3.5%  butter  fat  to  milk  con- 
taining 4%  butter  fat,  what  per  cent  of  the  whole  amount  of  milk  is 
the  skim  milk  removed? 

7.  How  much  each  of  6%  milk  and  skim  milk  should  be  mixed  to 
obtam  120  pounds  of  4%  milk? 

8.  What  are  the  proportions,  and  what  are  the  amounts,  of  skim 
milk  and  of  5%  milk  required  to  obtain  50  pounds  of  3%  milk? 

9.  How  many  pounds  of  5.2%  milk  must  be  added  to  200  lb.  of 
3%  milk  to  make  4%  milk? 

Solution  :   Let  a;  =  the  number  of  pounds  of  5.2%  milk  to  be  added. 

rp-L 

.04(200+0;)  =  .03  X  200+.052  x. 
Solve  this  equation  for  x. 

10.   How  many  pounds  of  3.1%  milk  must  be  added  to  54  lb.  of 
6.3%  milk  to  make  4%  milk? 


X,  §  144]  AVERAGES  AND  MIXTURES  169 

11.  Find  the  proportions  in  which  to  mix  5%  milk  with  3.5%  milk 
to  make  4%  milk.  How  many  pounds  of  each  should  be  taken  to 
make  90  pounds  of  4%  milk? 

12.  How  much  5.2%  milk  and  skim  milk  containing  .1%  butter 
fat  should  be  used  to  make  50  lb.  of  3.9%  milk? 

13.  How  many  pounds  of  .1%  skim  milk  should  be  added  to  5 
gallons  of  5.3%  milk  to  produce  3.5%  milk? 

14.  How  much  .1%  skim  milk  must  be  removed  from  780  lb.  of 
3%  milk  to  raise  the  butter  fat  content  to  4%? 

16.  I  wish  to  standardize  26%  cream  to  20%  cream  by  adding  4% 
milk.     What  are  the  proportions  of  cream  and  milk  to  be  used? 

16.  How  much  30%  cream  should  be  added  to  500  lb.  of  3%  milk 
to  make  5%  milk? 

17.  What  is  the  price  per  pound  of  butter  fat  when  20%  cream  sells 
for  150^  per  gallon? 

144.  Arithmetic  Mean,  Geometric  Mean,  Average.  The 
arithmetic  mean,  or  simple  average,  of  several  numbers  is 
their  sum  divided  by  their  number.  Thus  the  arithmetic 
mean  or  average  of  4  and  9  is  (4+9)/2=6|,  of  a  and  h  is 
(aH-6)/2,  of  a,  b,  c  is  {a-\-b-\-c)/3.  The  arithmetic  mean 
of  the  n  numbers  ai,  a2,  a^,    ••,  On  is 

(I)  j_gi+Q2+fl3+   •••  +gn. 

n 

For  example,  the  average  length  of  nine  ears  of  com  whose  measure-- 
ments  are  3",  4",  4",  6",  7",  7",  8",  9'',  10"  is 

^  ^  3+4+4+6+7+7+8+9+10  ^q^„ 

The  geometric  mean  of  two  numbers  is  the  square  root  of 
their  product ;  of  three  numbers,  is  the  cube  root  of  their 
product.  The  geometric  mean  of  n  numbers  ai,  02,  03,  -",  dn 
is  the  nth  root  of  their  product,  i.e. 


(2)  G=  Vfli  •02-03  •••  On. 

For  example,  the  geometric  mean  of  4  and  9  is  V4x9=c;  the 
geometric  mean  of  4,  9,  12  is  v/4x9x  12=7.56 


170  MATHEMATICS  \X,  §  144 

The  geometric  mean  of  two  numbers  is  very  often  called 
jhe  mean  proportional  between  the  two  numbers. 

An  average,  or  a  mean,  is  any  representative  type  which  is 
used  to  describe  a  whole  collection  of  values  or  objects. 
Thus  the  collection  of  nine  ears  of  corn  noted  above  is 
described  by  saying  that  the  average  length  of  ear  is  6|-". 

EXERCISES 

1.  A  man  walks  a  mile  at  the  rate  of  5  mi.  per  hour  and  a  second 
mile  at  the  rate  of  3  mi.  per  hour.  To  what  average  rate  for  the  two 
miles  is  this  equivalent  ? 

[Hint.  Average  time  per  mile  =  |  {\+l)  hr.  =-s\  hr.  This  is  3|  mi. 
per  hour.] 

2.  What  is  the  average  rate  for  two  miles  the  first  of  which  is  traveled 
at  the  rate  of  a  miles  per  hour  and  the  second  at  the  rate  of  b  miles  per 
hour?  ^^^      2ab_^ 

a-\-b 

3.  What  is  the  average  rate  for  three  successive  miles  traveled  at 
the  rates  of  15,  20,  25  mi.  per  hour?  Ans.     19.+  mi.  per  hour. 

4.  What  is  the  average  rate  for  four  successive  miles,  the  rates  for 
which  are  15,  20,  25,  30  mi.,  respectively?    Ans.     21.+  mi.  per  hour. 

5.  A  farmer  makes  8%  on  his  investment  the  first  year.  Adding 
this  to  his  capital,  he  makes  10%  on  the  amount  the  second  year,  and 
he  makes  15%  the  third  year.  What  is  the  average  rate  of  increase  for 
the  three  years? 

[Hint.  Let  r  be  the  average  rate  and  let  p  be  the  principal  he  starts 
with.  Then  the  amount  of  p  dollars  at  an  average  rate  r  for  3  years 
is  p(l+r)3.  This  must  equal  the  actual  amount  he  has  at  the  end  of 
3  years.     This  equahty  may  then  be  solved  for  r.] 

6.  What  is  the  average  rate  of  increase  r  of  a  certain  capital  if  the 
annual  rates  are  n,  r^,  Vz,  respectively  ? 

145.   Meaning  and  Use  of  an  Average  or  Mean.    We 

saw  in  §  144  that  any  representative  type  or  individual  is 
an  average  or  mean  in  so  far  as  it  represents,  in  some  way, 
the  whole  collection  to  which  it  belongs.     When  we  are 


X,  §  146]  AVERAGES  AND  MIXTURES  171 

concerned  with  groups  of  individuals  we  can  seldom  make  a 
definite  quantitative  statement  that  surely  applies  to  any 
given  individual.  But  we  can  make  statements  that  apply 
to  the  average  of  the  group  with  a  certainty  that  increases 
as  the  number  of  individuals  increases.  For  example,  the 
average  healthy  person  30  years  old  will  live  to  be  67,  though 
any  given  one  of  them  may  die  sooner  or  later  than  that. 
The  whole  business  of  life  insurance  is  based  upon  expecta- 
tions of  this  kind,  which  are  based  upon  the  experience  of 
life  insurance  companies  with  hundreds  of  thousands  of 
people.  Again,  we  could  not  be  certain  of  the  true  char- 
acteristics of  Jersey  cows  from  a  single  cow  or  from  a  few  of 
them,  but  after  a  study  of  measurable  characters  of  a  large 
group  of  them,  very  definite  statements  can  be  made. 

To  describe  in  a  few  words  or  symbols  a  whole  group 
whose  individuals  we  have  measured,  we  put  aside  the 
measurements  of  any  one  individual  and  construct  a  sin- 
gle, intermediate  number  or  characteristic.  This  gives 
some  kind  of  average,  which  will  be  descriptive  of  the 
whole  group. 

146.  Different  Kinds  of  Averages.  There  are  five  dif- 
ferent kinds  of  averages  in  common  use  for  different  pur- 
poses, viz.  : 

(1)  the  arithmetic  mean  (§  144), 

(2)  the  weighted  arithmetic  mean  (§  147), 

(3)  the  geometric  mean  (§  144), 

(4)  the  mode  (§  150), 

(5)  the  median  (§  149). 

All  of  these  are  in  general  practical  use.  Such  phrases 
as  average  rainfall,  average  rate  of  interest,  average  wage, 
average  college  student,  all  refer  to  different  kinds  of 
averages. 


172 


MATHEMATICS 


[X,  §  147 


147.  The  Weighted  Arithmetic  Mean.  A  slight  modifi- 
cation of  the  simple  arithmetic  mean  is  the  weighted  arith- 
metic mean.  For  example,  the  arithmetic  mean  of  the 
lengths  of  a  hundred  ears  of  corn  is  merely  the  sum  of  these 
lengths  divided  by  one  hundred.  However,  these  hundred 
measurements  may  be  arranged  in  half-inch  groups  as 
follows. 


Inches 

3.0 

3.5 

4.0 

4.5 

6.0 

5.5 

6.0 

6.6 

7.0 
8 

7.5 
10 

8.0 
10 

8.. 
14 

9.0 
13 

9.6 
10 

10.0 

10.5 
2 

n.o 

1 

Frequencies  or  number  of  ears 

■ 

2 

2 

2 

3 

4 

6 

7 

The  weighted  arithmetic  mean  of  these  lengths  is  found 
by  taking  each  group  length  times  the  number  in  the  group. 
The  sum  of  these  products  divided  by  the  total  number  of 
ears  is  the  average  length,  i.e. 

Av.  length 

_  1X3+2X3.5+2X4+  »•♦  +13X9+10X9.5+  •»  +2X10.5+1X11.0 

100 

=  7.7" 

In  general,  if  Wi,  m2,  rris,  •••,  m^  represent  the  class  marks 
{length  of  ear,  temperature,  per  cent  butter  fat,  etc.)  and  if 
Wi,  W2,  Ws,  •••,  Wr  the  frequencies  {number  of  ears,  gallons, 
pounds,  etc.),  then 

Wrmr  _ 

•Wr 


(3)     Weighted  Arithmetic  Mean=  ^^^^+^^^^+  ' 

^^1  +  ^2  +  ^3  + 


Thus,  the  weighted  arithmetic  mean  is  obtained  by  multiply- 
ing each  mark  by  the  corresponding  frequency,  and  dividing 
the  sum  of  these  products  by  the  total  number  in  the  collection. 

The  principal  advantage  of  the  weighted  arithmetic  mean 
over  the  simple  average  is  that  it  is  more  easily  computed. 

The  weighted  arithmetic  mean  is  widely  used  in  statistics, 
in  mixing  various  compounds  of  different  purities  or  of 
different  temperatures,  in  determining  results  of  civil  serv- 


X,  §  148]  AVERAGES  AND  MIXTURES  173 

ice  examinations  where  each  subject  is  weighted,  in  mixing 
concrete  from  substances  of  known  voids  as  sand  and  gravel, 
in  locating  centers  of  gravity  of  masses  Wi,  w^,  Ws,  •••,  Wf 
lying  along  a  straight  line  at  distances  Wi,  W2,  •••,  w,.,  re- 
spectively, from  a  fixed  point  in  that  line  (or  lying  in  a 
plane  at  these  distances  from  a  fixed  line),  etc.,  etc. 

Example.  Where,  on  a  uniform  bar  whose  own  weight  is  neglected, 
is  the  center  of  gravity  (balancing  point)  of  weights  of  1,  2,  3,  2,  2,  1 
pounds,  if  they  are  placed  at  distances  of  4",  6",  11",  12'',  20",  30", 
respectively,  from  one  end  of  the  bar? 

The  required  distance  from  that  end  is  the  weighted  average 

A  =  1X4+2X6+3  X11+2X12+2X20  +  1X30  _^3/, 

1+2+3+2+2+1 

148.  Weighted  Average  of  Grades  of  Milk.  It  is  fre- 
quently necessary  to  know  quickly  the  results  of  mixing 
grades  of  milk,  or  other  substances  of  different  purities  or  of 
different  temperatures.  Such  problems  may  be  stated  in  terms 
of  weighted  averages,  as  illustrated  by  the  following  example. 

Example.  What  kind  of  milk  is  a  mixture  of  3  gal.  of  5.3%  milk 
and  5  gal.  of  3.1%  milk? 

This  is  a  problem  in  simple  averages  (§  144)  if  we  allow  the  repetition 
of  the  5.3%  grade  3  times  and  of  the  3.1%  grade  5  times,  i.e. 

Average  =  (5.3+5.3+5.3)  +  (3.1  +3.1+3.1+3.1  +3.1) 

^3(5,^+5(M)  =.3.9%  milk. 
3+5 

Thus,  generally,  Wi  pounds  of  an  mi  milk  and  W2  pounds  of  an  mi 
milk  will  give  an  average  milk 

Wi-\-W2 

Clearing  (4)  of  fractions  and  collecting,  we  have 

Wi(A  —mi)=W2(m2  —  A), 
or 

(5)  "!£!  _  1^2— A 

W2~  A  —mi 


174 


MATHEMATICS 


p:,  §  148 


Equation  (5)  gives  the  ratio  of  the  amounts  of  milk  in  terms  of  the 
kinds  of  milk.  This  leads  at  once  to  the  so-called  parallelogram  method 
of  standardizing  mixtures.  This  is  merely  a  graphic  form  of  interpreting 
equation  (5)  as  follows. 

Place  the  average  in  the  center  of  the  parallelogram,  Fig.  93,  and  the 
marks  of  the  substances  concerned  {per  cent,  strength,  temperature,  etc.), 
at  the  left-hand  corners.  Subtract  diagonally,  placing  the  remainders 
{taken  positively)  at  the  opposite  right-hand  corners  respectively.  These 
remainders  are  the  ratios  of  the  substances  mixed  which  produce  the  given 
or  the  desired  average. 

'A 


Fig.  93.  —  Dairymen's  Parallelogram 


Example  1 
should  be  added  to  produce  a  3.5%  milk? 

Let  x=  the  number  of  pounds  of  5%  milk  to  be  added, 
from  the  diagram 


How  much  5%  milk 


Therefore. 


6^ 


.5^- 


-*-xlb. 


j|J!^  =  j|,  from  which 

a:  =  50  1b. 

Example  2.     A  farmer  receives  an  order  for  50  lb.  of  3.6%  milk. 

How  many  pounds  each  of  3.1%  milk  and  4.7%  milk  shall  he  take  to 

fill  the  order? 

Let  x  =  the  number  of  pounds  of  3.1%  milk  required  and 
50— a:  =  the  number  of  pounds  of  4.7%  milk  required. 

Therefore,  from  the  diagram 

8.1  p^^ ^-\  1.1 >  X  lb. 


we  have 


4.7 


->50-a?». 


=  -^ ,  from  which 


50-a:       .5 
a;  =  34.4  lb 


X,  §  148] 


AVERAGES  AND  MIXTURES 


175 


EXERCISES 

1.  A  farmer  mixes  8  gal.  of  3%  milk,  10  gal.  of  2.9%  milk,  5  gal.  of 
3.5%  milk,  4  gal.  of  5.3%  milk,  6  gal.  of  3.7%  milk.  What  per  cent 
butter  fat  is  the  mixture? 

2.  What  kind  of  milk  must  be  added  to  the  mixture  of  Ex.  1  to  make 
36gal.  of  3.5%milk? 

3.  How  much  water  must  be  added  to  a  mixture  of  3  pt.  of  96% 
alcohol  and  8  pt.  of  78%  alcohol  to  make  the  final  mixture  60%  alcohol? 

[Hint.     Let  x  =  the  amount  of  water  to  be  added.] 

4.  Masses  of  4,  7,  9,  13  pounds  are  placed  on  a  straight  hne  at 
distances  of  +2',  —3',  —4',  +5',  respectively,  from  a  given  point  0  in 
that  lijie.     Where  is  the  center  of  gravity  with  respect  to  this  point? 

Ans.     +3!'  from  the  point. 

6.  Weights  of  1,  2,  3,  4  pounds,  respectively,  are  placed  in  order 
at  the  corners  of  a  square  1'  on  each  side.  Find  the  center  of  gravity 
of  this  square,  neglecting  the  weight  of  the  square  itself. 

[Hint.  Let  two  sides  of  the  square  intersecting  at  the  one  pound 
weight  be  the  x-  and  y-axes  as  in  the  accompanying  figure.  Find  the 
x-distance  of  the  center  of  gravity  as  in  the  pre- 
vious exercise.     Similarly  for  the  i/-distance.] 

6.  Find  the  center  of  gravity  of  an  equi- 
lateral triangle  on  which  weights  of  1, 2, 3  pounds, 
respectively,  are  placed  at  each  corner. 

7.  A  candidate  receives  the  following  marks 
in  a  civil  service  examination.  In  penmanship, 
weighted  1, 90% ;  in  geography,  weighted  2, 75% ;  in  arithmetic,  weighted 
5,  70% ;  in  U.  S.  History,  weighted  3,  60% ;  in  English,  weighted  2, 
80%.  Compare  his  standing  with  the  candidate  who  has,  for  the  same 
subjects  and  weights,  marks  of  85%  in  penmanship,  77%  in  geography, 
60%  in  arithmetic,  70%  in  U.  S.  History,  and  70%  in  English. 

[Hint.  A  weight  of  1  in  penmanship  and  of  2  in  geography  mean 
merely  that  geography  is  rated  as  twice  as  important  as  penmanship.] 

8.  Find  the  weighted  arithmetic  mean  for  the  following  table  of 
measurements  of  54  seeds. 


2  lb 

Y 

3  lb. 

0 

1  lb. 

4 

lb. 

Fig.  94 


Length  in  mm.      .    .    . 

1.56 

1.58 

1.59 

1.60 

1.61 

1.62 

1.63 

1.64 

1.65 

1.66 

1.67 

1.68 

Number  of  Seeds .     .     . 

1 

' 

1 

2 

2 

6 

11 

13 

7 

4 

5 

2 

9.  Solve  Exs.  7-16,  §  143,  by  the  parallelogram  method. 


176  MATHEMATICS  [X,  §  149 

149.  The  Median.  The  median  of  a  collection  is  a  value 
or  type  such  that  half  of  the  other  values  exceed  it  and  half 
are  below  it.  If  the  numbers  are  arranged  in  numerical 
order  in  a  column,  the  number  that  is  halfway  down  the 
column  is  the  median.  It  may  be  found  also  by  pairing 
off  the  largest  and  the  smallest  and  repeating  the  process 
until  only  one  or  two  is  left.  If  two  are  left,  the  median 
may  then  be  taken  halfway  between  them. 

Example  1.     The  median  of  1,  3,  3,  5,  6,  8,  10  is  5. 

Example  2.  The  median  wage  of  825  workmen  is  that  of  the 
middlemost  of  the  men  when  they  are  arranged  in  numerical  order 
with  respect  to  wages,  i.e.  if  412  receive  more  than  $2.77  and  412  re- 
ceive less,  then  $2.77  is  the  median  wage. 

The  advantage  of  the  median  as  a  representative  type  is 
the  fact  that  it  can  be  determined  easily.  Very  large  or 
very  small  values  do  not  affect  it.  However,  it  gives  no 
special  importance  to  extreme  values  and  may  be  after  all 
far  removed  from  the  type. 

150.  The  Mode.  The  mode  is  the  most  frequent  value. 
In  the  set  of  values  3,  4,  4,  5,  6,  the  mode  is  4  since  it  occurs 
twice  and  the  others  only  once. 

In  a  popular  way,  mode  means  fashion.  Thus  that  type 
of  citizen  met  oftener  than  any  other  is  an  average  citizen. 
For  a  community  that  has  three  millionaires  but  all  the 
other  citizens  in  poverty,  the  arithmetic  average  for  the 
financial  condition  of  the  community  would  give  the  impres- 
sion that  all  the  citizens  were  well-to-do,  while  in  reality  the 
average  citizen  is  in  poverty.  The  mode  is  useful  in  describing 
such  a  situation. 

EXERCISES 

1.  Find  the  median  length  for  the  100  ears  of  corn  in  the  table  of  §  147. 

2.  Find  the  median  length  of  seed  for  the  seeds  tabulated  in  Ex.  8, 
§148. 


X,  §  151]  AVERAGES  AND  MIXTURES  177 

3.  Which  mean  is  meant  in  the  following  cases?  Average-sized 
apple;  average  coloration  of  Rome  Beauty  apples;  average  price  of 
butter;  average  wheat  production  per  acre  in  the  United  States; 
average  rate  of  depreciation  of  farm  machinery;  mean  daily  tempera- 
ture ;  normal  rainfall ;  average  student. 

4.  What  per  cent  increase  per  year  will  double  the  attendance  at  a 
University  in  10  years'  time? 

[Hint.     Use  v^2.     Why?     Compute  this  value  by  logarithms.] 

5.  What  is  the  average  annual  rate  of  increase  of  wheat  production, 
if  it  has  increased  from  525  millions  of  bushels  to  775  millions  in  10 
years'  time? 

6.  What  kind  of  averaged  age  is  it,  if  out  of  a  total  of  100,000  people 
alive  at  age  10,  about  one  half,  viz.  51,230,  are  living  at  age  64? 

7.  What  is  the  mode  for  the  table  of  measurements  in  Ex.  8,  §  148? 

151.  Mixing  Fertilizers.  The  use  of  commercial  ferti- 
lizers is  one  method  of  returning  to  the  soil  plant  foods  needed 
by  growing  plants.  These  fertilizers  contain  nitrogen, 
phosphoric  acid,  and  potash  in  various  proportions  for 
different  crops  and  soils.  It  is  quite  often  more  satisfactory 
and  much  cheaper  to  buy  the  fertilizing  materials  and  mix 
them  on  the  farm. 

A  fertilizer  labeled  2-8-4  contains  2%  nitrogen,  8%  phos- 
phoric acid,  and  4%  potash.  Nitrogen  is  obtained  chiefly 
from  nitrate  of  soda,  dried  blood,  dried  fish  scrap,  and 
cottonseed  meal.  Phosphoric  acid  (phosphorus)  is  obtained 
from  ground  bone,  basic  slag,  ground  phosphate  rock,  and 
acid  phosphate.  Potash  is  obtained  from  muriate  of 
potash,  sulphate  of  potash,  and  kainite.  The  usual  compo- 
sition of  fertilizing  materials  is  about  as  follows.  Chilean 
nitrate  of  soda  contains  15%  nitrogen ;  acid  phosphate 
contains  about  16%  phosphoric  acid ;  muriate  of  potash 
contains  about  50%  potash.  More  complete  tables  of  fer- 
tilizing materials  will  be  found  in  the  Appendix,  Tables  III 
and  IV. 


178  MATHEMATICS  \X,  §  151 

Filler,  usually  in  the  form  of  sand  or  dry  dirt,  is  added  to 
make  the  proper  per  cent  and  to  aid  in  distributing  the 
fertilizer  evenly. 

EXERCISES 

1.  A  farmer  wishes  to  make  1  ton  of  2-8-4  fertilizer  from  Chilean 
nitrate,  acid  phosphate,  and  muriate  of  potash.  (See  §  151.)  How 
much  of  each  ingredient  must  he  use  and  how  much  filler  must  he  add  ? 

[Hint.  The  nitrogen  content  is  2%  of  2000  pounds  =  40  pounds. 
This  must  be  obtained  from  Chilean  nitrate  which  is  only  15%  nitro- 
gen. Therefore  if  x  is  the  number  of  pounds  of  nitrate  used,  .15  x  =  40 ; 
hence  x  =  267  pounds.]  Ans.     Filler  =  573  pounds. 

2.  Show  how  to  make  one  ton  of  3-10-3  fertilizer,  which  is  used  for 
raising  corn  or  wheat.  Calculate  its  cost  at  3f^  per  pound  for  Chilean 
nitrate,  1^  per  pound  for  acid  phosphate,  and  3^  per  pound  for  muriate 
of  potash. 

3.  Show  how  to  make  a  4-9-4  fertilizer,  used  for  raising  potatoes, 
from  the  materials  of  Ex.  2.     Calculate  the  cost  per  ton. 

4.  A  farmer  has  a  ton  of  fertilizer  containing  30%  dried  blood, 
62 f%  acid  phosphate,  and  7|%  muriate  of  potash.  How  many  pounds 
of  acid  phosphate  must  he  add  to  this  ton  to  raise  the  acid  phosphate 
content  of  the  mixture  to  70%?  Ans.     500  1b. 

5.  Compare  a  2-8-2  fertilizer  at  $25.00  per  ton  and  a  2-8-6  fertilizer 
at  $30.00  per  ton  on  the  basis  of  the  average  cost  per  pound  of  plant 
food  which  they  contain.     What  kind  of  average  is  this? 

6.  How  many  pounds  of  acid  phosphate  must  be  added  to  400  lb. 
of  a  2-6-4  fertilizer  to  convert  it  into  a  2-8-4  fertilizer,  if  the  acid  phos- 
phate contains  16%  phosphoric  acid?  Ans.     100  pounds. 

7.  How  many  pounds  of  acid  phosphate  and  how  many  of  muriate 
of  potash  must  be  added  to  a  ton  of  a  2-6-4  fertilizer  to  convert  it  into 
a2-8-6fertiUzer? 

8.  A  farmer  mixes  500  pounds  of  a  2-9-5  fertiHzer  and  300  pounds 
of  a  2-6-2  fertilizer.  What  ratios  do  the  ingredients  of  the  mixture 
have? 

9.  Find  the  ratios  of  the  ingredients  for  200  lb.  of  a  2-6-4  fertilizer 
mixed  with  300  lb.  of  a  1-2-1  fertilizer. 


X,  §  151]  AVERAGES  AND  MIXTURES  179 

In  solving  the  following  problems,  use  Tables  III  and  IV,  Appendix, 
which  give  the  required  amounts  of  fertilizing  materials  per  acre  to 
grow  various  crops,  and  which  give  the  composition  of  these  materials. 

10.  What  quantities  of  dried  blood  (13%  nitrogen)  and  bone  meal 
(3%  nitrogen,  24%  phosphate)  are  needed  for  an  acre  of  wheat  ground 
which  is  strong  in  potash  but  which  requires  12  lb.  nitrogen  and  20  lb. 
phosphate  per  acre  ? 

Solution  :    Let         x  =  \h.  dried  blood  required. 
2/  =  lb.  bone  meal  required. 

In  X  lb.  dried  blood  there  are  .13  x  lb.  nitrogen,  and  in  y  lb.  bone  there 
are  .03  y  lb.  nitrogen  and  .24  y  lb.  phosphate.  Hence  the  total  nitrogen  is 
.13  X+.03  y  =  l2  lb.,  and  the  total  phosphate  is  24  2/  =  20  lb.  Solving 
these  two  equations  for  x  and  y,  we  find  x  =  73  lb.,  y  =  83  lb. 

11.  In  Ex.  10,  substitute  nitrate  of  soda  and  phosphate  of  lime 
for  dried  blood  and  bone  meal. 

12.  In  Ex.  10,  substitute  com  for  wheat. 

13.  A  gardener  has  a  supply  of  wood  ashes  and  a  limited  amount 
of  hen  manure  mixed  with  an  equal  weight  of  leaves.  How  much 
dissolved  bone  should  he  buy  to  make  a  complete  fertilizer  for  an 
acre  of  beets? 

Solution  :    Analysis  of  hen  manure  2      —     2      —  1      % 

Analysis  of  leaves  .7   —       .15—    .3  % 

Average  for  equal  amounts  1.35—  1.08—  .65% 
Analysis  of  dissolved  bone  2  —  20  —  0  % 
Analysis  of  wood  ashes  0      —     1      —  5      % 

The  requirement  for  an  acre  of  beets  is 

20  lb.  nitrogen  — 25  lb.  phosphate— 35  lb.  potash. 
Let  x  =  no.  of  lb.  of  mixed  hen  manure  and  leaves  required  per  acre, 
2/  =  no.  of  lb.  of  wood  ashes  per  acre, 
2  =  no.  of  lb.  of  dissolved  bone  per  acre. 
Then  the  total  nitrogen  is 

.0135  a;  +  .02  z  =  20  lb., 

the  total  phosphate  is 

.0108  x  +  .Ol  2/ +  .20  2  =  25  lb., 
and  the  total  potash  is 

.0065  x  +  . 05  2/  =35  lb. 

Solving  these  equations,  we  find  approximately, 
x  =  1460  1b.,  2/  =  510  1b.,  2  =  21  lb. 


180  MATHEMATICS  pC,  §  151 

14.  A  man  wishes  to  raise  an  acre  of  potatoes.  He  has  at  hand 
500  lb.  farmyard  manm^e  and  100  lb.  hen  manure.  These  are  to  be 
mixed  in  this  proportion  when  used  with  wood  ashes  to  make  a  minimum 
complete  fertilizer  for  the  particular  soil  used.  How  many  pounds  of 
each  are  required? 

15.  Find  the  proportions  of  wood  ashes  and  hen  manure  for  an  acre 
of  celery  on  soil  that  requires  no  additional  nitrogen. 

16.  A  man  has  applied  4000  lb.  of  farmyard  manure  to  an  acre  of 
tomato  land.  How  much  wood  ashes,  dried  ferns,  and  hen  manure, 
mixed  in  the  proportions  4,  4,  and  1,  respectively,  should  he  use  to 
complete  the  f ertihzation  ? 

17.  Find  a  complete  fertiUzer  made  of  acid  bone  and  sulphate  of 
ammonia  for  an  acre  of  cabbage  on  clay  land  that  requires  no  potash. 

18.  What  quantities  of  floats,  farmyard  manure,  and  sulphate  of 
potash  are  needed  for  raising  an  acre  of  onions  ? 

152.  The  Mixing  of  Concrete.  Concrete  is  usually  a 
mixture  of  cement,  sand,  and  gravel  (c,  s,  g),  in  various  pro- 
portions, depending  upon  the  use  to  which  it  is  to  be  put. 
Since  there  is  a  certain  per  cent  of  air  space  {voids)  in  sand 
and  gravel,  it  is  necessary  to  calculate  the  various  amounts 
of  c,  s,  g  to  make  a  given  volume,  say  1  cu.  yd.,  of  solid  con- 
crete. A  common  working  rule  is  to  allow  33^%  voids  in 
the  sand  and  45%  voids  in  the  gravel  or  crushed  stone. 

Example.     How  many  cubic  yards  each  of  cement,   sand,   and 
gravel,  mixed  in  the  ratio  1-2-4  (i.e.  1  part  cement  to  2  parts  sand  to 
4  parts  gravel)  are  required  to  make  one  cubic  yard  of  solid  concrete, 
allowing  33^%  voids  in  the  sand  and  45%  voids  in  the  gravel? 
Let  X  =  the  cubic  yards  cement, 

y  —  the  cubic  yards  sand, 
0  =  the  cubic  yards  gravel  required. 

Since  the  voids  in  the  gravel  are  filled  up  in  the  finished  concrete, 
the  z  cu.  yd.  of  gravel  count  as  only  .55  z  cu.  yd.  of  solid  concrete,  and 
the  y  cu.  yd.  of  sand  count  as  only  |  y  cu.  yd.  of  solid,  finished  concrete. 
There  being  no  voids  in  the  cement,  we  have  the  equations : 

(1)  x+f  ?/  +  .55  z  =  1  (cu.  yd.  of  concrete) ; 


X,  §  153] 


AVERAGES  AND  MIXTURES 


181 


(2)  y  =  2x  and  z=^4x  from  the  ratios  of  the  mixture. 
Substituting  the  values  of  y  and  z  from  (2)  in  (1),  we  have 

a:  +  ^-f2.2x  =  l. 

Solving  this  equation  we  obtain 

X  =  .22  cu.  yd.  cement, 
from  which  we  find  y  =  .4A  cu.  yd.  sand,  and  z  =  .88  cu.  yd.  gravel. 

153.  Unit  of  Volume  of  Cement.  Cement  varies  greatly 
in  weight  in  different  stages  of  compactness.  Results  show 
a  variation  from  86  to  123  pounds  in  the  average  weights 
of  the  same  cement  according  as  it  was  weighed  sifted  or 
packed  in  a  barrel.  The  quantity  of  cement,  in  the  pro- 
portion for  mixing  concrete,  should  therefore  invariably 
be  regulated  by  its  weight.  In  case  the  proportions  are  by 
volume,  then  a  definite  weight  or  a  definite  number  of 
packages  of  cement  must  be  assumed  as  the  unit  volumes. 

As  to  the  unit  which  should  be  selected  for  the  volume  of  a 
cement  barrel,  the  opinion  of  some  fifty  authorities  (mem- 
bers of  the  American  Society  of  Civil  Engineers)  differed 
greatly.  An  approximate  average  of  all  the  figures  sug- 
gested by  them  was  that  a  barrel  of  cement  should  be 
3.8(3.76)  cubic  feet,  which  corresponds  to  100  pounds  to 
the  cubic  foot.     There  are  four  sacks  of  cement  in  1  barrel. 

EXERCISES 
Complete  the  following  table  giving  the  ingredients  for  1  cu.  yd.  of 
solid  concrete,  allowing  33^  %  voids  in  the  sand,  40  %  in  the  gravel. 


NintfBER 

Mixture 

Cement 

Sand 

Gravel 

Cu.  Yd. 

Sacks 

1 

1-2-3 

7.2 

2 

1-21-5 

3 

1-3-5 

4 

1-3-6 

5 

1-2-2 

182  MATHEMATICS  [X,  §  153 

6.  Give  the  ingredients  for  a  cubic  yard  of  1-2-3  concrete  allowing 
20%  voids  in  the  sand  and  40%  in  the  crushed  stone. 

7.  Solve  as  in  Ex.  6  for  a  1-3-5  concrete. 

154.  Proportions  of  Cement,  Sand,  Gravel  for  Various 
Purposes.  The  following  are  the  most  commonly  accepted 
proportions  of  cement,  sand,  and  gravel,  respectively,  for 
various  purposes. 

(a)  Rich :  l-l|-3  for  columns  and  structural  parts  under 
very  heavy  stress. 

(6)  Standard  :  1-2-4  for  floors,  beams,  and  columns  under 
considerable  stress. 

(c)  Medium :  l-2|-5  for  walls,  piers,  sidewalks,  etc.,  and 
not  subject  to  great  stress. 

(d)  Lean  :  1-3-6  for  heavy  mass  work  under  compression 
only. 

155.  Fuller's  Rule.  For  determining  more  quickly  the 
amount  of  the  ingredients  in  a  cubic  yard  of  concrete,  the 
following  formulas,  known  as  Fuller's  Rule,  give  rather  good 
working  results : 

(5)  C= =  the  barrels  of  cement  in   1   cu.  yd.  of 

concrete, 

(6)  S=  ^^^^^^-^^  =  the  cubic  yards  of  sand  in  1  cu.  yd.  of 

concrete, 

(7)  G'=g(g)=:^fa)(3-8)^the  cubic   yards   of    gravel  in 

1  cu.  vd.  of  concrete, 
where 

c  =  the  number  of  parts  of  cement  =  1  always, 

s  =  the  number  of  parts  of  sand, 

Sr  =  the  number  of  parts  of  gravel. 

To  reduce  barrels  of  cement  to  sacks,  multiply  the  number 
of  barrels  by  4. 


X,  §  155] 


AVERAGES  AND  MIXTURES 


183 


EXERCISES 

Fill  out  the  accompanying  diagram  according  to  Fuller's  Rule  and 
thus  obtain  some  working  rules. 


Parts  in  the  FoRMtrLA 


1.  l-U-3 

2.  1-2-4 

3.  1-21-5 

4.  1-3-6 


Sacks  Cement 


Cu.  Yd.  Cement 


Cu.  Yd.  Gravel 


5.  Compare  the  amounts  of  material  necessary  for  1  cu.  yd.  of  a 
1-2^-5  concrete  mixed  according  to  Fuller's  Rule  and  according  to  the 
example  of  §  152. 


CHAPTER  XI 


GEOMETRY  —  MENSURATION 

156.  Right  Triangles.  Some  elementary  mensuration 
rules  were  stated  in  Chapter  II.  The  purpose  of  the  present 
chapter  is  to  simplify  what  has  been  given,  and  to  give 
geometric  rules  and  theorems  of  more  advanced  character. 

In  a  right  triangle  ABC,  shown  in  Fig.  95,  let  the  sides 
be  denoted  by  a,  h,  c,  in  the  usual  manner  (§  15),  and  let 
the  perpendicular  h,  let  fall  from  C  on  the  side  c,  divide 
c  into  the  two  parts  m  and  n  (Fig.  c 

95).     Then  the  following  relations 
hold: 

(1)  a'^  +  b'  =  c\ 

(2)  /i2  =  mn, 


(3) 


mc,  b^ 


Fig.  95 


nc. 


Each  of  these  theorems  may  be  stated  in  words  by  the 
student :  (1)  is  the  Pythagoras  Theorem  (§  15) ;  (2)  and 
(3)  may  be  obtained  easily  from  the  fact  that  the  triangles 
ABC,  ADC,  and  BDC  are  all 
similar ;  (1)  is  obtained  by  adding 
the  two  parts  of  (3). 

157.   Oblique        Triangles.     In 

any  triangle  ABC  (Fig.  96),  the 
following  theorems  hold. 

An  exterior  angle  made  by  any 
side  and  an  adjacent  side  produced  is  equal  to  the  sum  of 
the  two  opposite  interior  angles. 

184 


XI,  §  158]         GEOMETRY  —  MENSURATION 


185 


The  altitudes  meet  in  a  point  0,  called  the  orthocenter. 

The  medians,  joining  each  vertex  with  the  midpoint  of 
the  opposite  side,  meet  in  the  center  of  gravity  G.  This 
point  G  trisects  each  median. 

The  perpendiculars  erected  at  the  midpoints  of  the  sides 
meet  in  a  point  Q  which  is  the  center  of  the  circumscribed 
circle. 

In  any  triangle  0,  G,  and  Q  lie  in  a  straight  line,  and  G 
is  two  thirds  of  the  way  from  0  to  Q. 

The  bisectors  of  the  angles  meet  in  a  point  P  which  is  the 
center  of  the  inscribed  circle  (Fig.  97). 
o 


c 
FiQ.  98 


The  bisector  of  an  angle  divides  the  opposite  side  into  seg- 
ments which  are  proportional  to  the  adjacent  sides  (Fig.  98). 


158.  Extension  of  the  P5rthagoras  Theorem  for  Any 
Triangle.  In  any  triangle  ABC  (Figs.  99,  100),  if  h  is  the 
altitude  on  b,  and  x  is  the  projection  of  c  on  b,  it  is  shown 


r — ^ ^. 

Fig.  99.— a2  =  62  +  c2-26x 


b  C 

Fig.  100.— o2  =  62  +  c2  +  26x 


in  plane  geometry,  by  using  the  Pythagoras  theorem  on  the 
two  right  triangles  ABD  and  DCB,  that 


186  MATHEMATICS  \XI,  §  158 

(4)  fl2  =  52_|_c2 _  2  bx,  if  A  is  acute, 
or 

(5)  a2=  &2+C2+2  bx,  if  A  is  obtuse. 

Solving  these  equations  for  x,  we  may  write  also 

2^2  _|_  ^2  _  q2 

(6)  X  =  — — ,  if  A  is  acute, 

2  o 

or 

(7)  X  =  g'-^'-g' ^  if  A  is  obtuse. 

2  u 


159.   Computation  of  Altitudes.    Area  of  a  Triangle.     If 

the  three  sides  of  a  triangle  are  given,  we  may  compute 

(a)  The  projections  of   any  side  upon   any  other  side 
(§  158). 


(6)  The  altitudes.  From  Fig.  99  or  100,  h  =  Vc'  -  x"^  where 
X  is  known  from  (6)  or  (7),  §  158. 

(c)  The  area.  When  the  altitude  has  been  found,  the 
area  follows  at  once.  The  formula  for  area  of  a  triangle 
from  §  15,  viz. : 


A  =  y/s{s  —  a){s  —  b)  (s—c), 
may  be  derived  from  §  158.     For,  from  Figs.  99,  100, 


V       26       J 


,2     ^2_^2    (b'+c''-ay_4¥c^-{¥+c''-ay, 

4  62  ' 


or,  factoring  the  numerator, 

;,2_  (a+6+c)(6+c-a)(q-6+c)(a+6-c) 
^"  462 

Let  a+6+c  =  2  s.    Then  we  have 
a+6-c  =  2s-2c  =  2  (s-c), 
a-6+c  =  2s-26  =  2  (s-6), 
-a+6+c=2  s-2  a  =  2  (s-a). 


XI,  §  160]        GEOMETRY  —  MENSURATION  187 

Therefore 

i,2_  2  g  2  (s-a)  2  (s-b)  2  js-c) 
^  4b^  ""' 


and 


(8)  h  =  Ws{s-a)(s-h){s-c). 

0 

Hence  the  area,  which  is  equal  to  hX-,is  also  equal  to 

(9)  Area  =  \/s{s  —  a){s—  b)  (s  —  c). 


EXERCISES 

1.  The  sides  of  a  triangle  are  9,  12,  15.  Find  the  segments  of  the 
longest  side  made  by  bisecting  the  angle  opposite  it. 

2.  The  sides  of  a  triangle  are  12,  14,  16.  Draw  it  to  scale  using  14 
as  the  base  and  find 

(a)  The  segments  of  the  side  14  made  by  the  bisector  of  the  opposite 
angle. 

(h)  The  projections  of  the  sides  12  and  16,  respectively,  on  the  side 
14. 

(c)  The  altitude  on  the  side  14. 

(d)  The  area  of  the  triangle. 

3.  Find  the  area  of  a  triangular  pasture  whose  sides  measure  19> 
32,  25  rods,  respectively. 

160.  The  Circle.  To  the  elementary  facts  about  circles 
given  in  §§  19,  20,  we  may  add  the  following  theorems,  which 
are  usually  proved  in  elementary 
geometry.  a^ 

I.  An  angle  between  two  secants  is 
measured  by  one  half  of  the  difference   d\ 
of  the  intercepted  arcs. 

Thus,  we  have  for  example  Z  A  PD  ^^^   ^^^ 

=  i  (arc  AD  -  arc  BC),  Fig.  101. 

II.  An  angle  between  two  chords  is  measured  by  one  half  of 
the  sum  of  the  intercepted  arcs. 

Thus,  ZEOF  =  i  (arc  EF +  a.rc  GH),  Fig.  101. 


188  MATHEMATICS  [XI,  §  160 

III.  If  a  tangent  PT  and  a  secant  PAB  he  drown  from 
any  exterior  point  P  to  a  circle  (Fig.  102),  the  square  of  the 
tangent  PT  is  equal  to  the  product 
of  the  tivo  segments  of  the  secant, 
PA  and  PB ;  i.e. 

(10)       PT'=PA  '  PB=PC'PD 

=  a  constant. 

This  fact  is  of  interest  with  refer- 
ence to  visibility  above  the  surface  of  the  earth.     Let  PB,  Fig. 
103,  be  a  diameter  of  the  earth,  PA  the  distance  above 
the  earth's  surface,  and  PT  the  distance  to  the  horizon. 
Then  p 

PA  •  PB  =  PT\ 
LetFA-fc;thenPB  =  /c+2r.   AlsoletPr=d 
Therefore  d''  =  k{k-\-2r). 

Practically,  h  is  small  as  compared  with  the 
diameter  of  the  earth.  Thus,  k-\-2r  may  be 
written  as  2  r.     Hence  ^^Q-  i^^ 

d^=2  kr,    or    k=  -— —  (k  and  d  in  miles). 
8000 

But  k  is  usually  measured  in  feet.    Therefore  let  k 

,  from  which 


5280 
and  7  ^     h    ^    d^ 

5280     8000^ 

h,  the  height  fn/eef  =  f|f  d'^=l  d"^,  approximately.* 

Therefore,  the  height  in  feet  equals  two  thirds  the  square  of 
the  distance  in  miles. 

Example.      A  lighthouse  is  240  feet  high.     How  far  is  its  hght 
visible  at  sea? 

SoiiUTioN :  We  have  h  =  240 ;  hence 

240  =  I  c/^    or    (i  =  19  miles. 

*  See  Exercise  9,  §  79. 


XI,  §  162]  GEOMETRY  —  MENSURATION 


189 


Fig.  105 


161.  Compound  Curves.  Arcs  AB  and  AC  of  two 
circles,  which  are  tangent  to  each  other  at  A,  Figs.  104-105, 
form  a  compound  curve.  The 
point  of  tangency  A  is  the  /^-=? 
transition  point  from  one  circle  ^ ' 
to  the  other. 

Compound  curves  are  used 
in    the    construction    of    rail- 
roads and  highways  where  the  track  must  be  made  to  con- 
form to  the  physical  features  of  the  land. 

162.  The  Area  of  a  Segment  of  a  Circle.  In  §  22,  mention 
was  made  of  the  area  of  a  segment  of  a  circle.  In  Fig.  106, 
let  h  =  height  of  a  segment, 

r  =  radius  of  the  circle. 
Then  OS  =  r  —  h,  and  the  angle  SOP  is  given  by 

(11)  cos  SOP  =  g|. 

The  area  of  the  sector  OPQR  and  of  the 
triangle  OPR  may  now  be  found.     The  required  area  of  the 
segment  is  their  difference. 


EXERCISES 

1.  Find  the  diameter  of  a  broken  wheel  a  chord  of  which  measures 
16"  and  the  height  of  the  corresponding  arc  is  4". 

2.  The  chord  of  an  arc  is  24"  and  the  chord  of  half  this  arc  is  13". 
Find  the  diameter  of  the  circle. 

3.  Show  that  the  following  construction,  Fig. 
107,  will  divide  a  circle  into  three  parts  whose 
areas  are  equal. 

Trisect  the  radius  at  B  and  C.     Draw  a  semi-  A\ 
circle  on  OA  as  diameter.     Erect  perpendiculars 
BR,  CP  meeting  the  semicircle  at  R,  P.     With 
centers  at  O,  draw  the  required  circles  with  radii 
OP  and  OR.  Fig.  107 


190 


MATHEMATICS 


[XI,  §  162 


4.  A  line  segment  approximately  equal  to  the  circumference  of  a 
circle  may  be  constructed  as  follows. 

Draw  a  diameter  AD,  Fig.  108,  and  COT  =  30°.  Draw  CT  perpen- 
dicular to  AD.  From  D  draw  DR  perpendicular  to  AD  and  equal  in 
length  to  three  times  the  diameter. 
Then  CR  is  approximately  equal 
to  the  circumference. 

Show  that  the  error  in  the  cir- 
cumference, constructed  in  this 
way,  is  about  1  in  20,000.  ^^^f.  108 

Use  the  formula  of  §  160  to  solve  the  following  exercises. 

5.  How  high  must  an  aeronaut  rise  above  the  earth  in  order  to  see 
a  city  50  miles  away?     How  high  in  order  to  see  100  mi. ? 

6.  How  far  can  a  man  6  ft.  tall  see  when  standing  on  a  level  plain? 

7.  The  lookout  in  the  rigging  of  a  vessel  is  100  feet  above  the  water. 
How  far  can  he  see  a  beacon  light  known  to  be  150  feet  high? 

8.  The  Woolworth  building  in  New  York  City  is  750  feet  high. 
How  far  can  it  be  seen  by  the  lookout  on  a 
vessel,  if  the  lookout  is  75  feet  above  the  water  ? 


9.   The  curve  ABC  of  a  railroad  track  is  com- 
posed of  two  circular  arcs  AB  and  BC,  as  in  the 
accompanying  figure.     The  center  0  of  BC  is  on 
the  radius  PB  of  AB.     Prove  that  ABC  is  a  compound  curve,  i.e.  that 
ABf  BC  are  arcs  of  tangent  circles. 

10.  The  track  of  a  switch  AOBC  from  a  straight  railroad  track 
AD,  as  in  the  accompanying  figure,  is  laid  out  as  follows.  Arc  AO  is 
constructed  with  center  M,  where  AM  is  per-  j^ 
pendicular  to  AD.  MO  is  produced  to  N  so 
that  MO=ON  and  arc  OB  with  center  N  is 
drawn. 

Show  that  AOB  is  a  compound  curve.  Fig.  110 

11.  By  sawing  off  4"  of  the  diameter  of  a  log  12"  in  diameter  what 
part  of  the  area  is  sawed  ? 

12.  The  height  of  a  segment  of  a  circle  is  3"  and  its  base  is  10". 
Find  the  radius  and  the  area  of  the  corresponding  circle,  and  the  area 


M 


of  the  segment. 


Ans. 


=  51". 


XI,  §  163]        GEOMETRY  —  MENSURATION 


191 


/        F'/ 

\                       ^ 

FiQ.  111.  —  Ellipse  with  Axis 
Ai«)  Foci 


163.  The  Ellipse.  An  ellipse  is  formed  when  a  point  moves  so 
that  the  sum  of  its  distances  from  two  fixed  points  ls  constant. 

The  two  fixed  points  are  called  foci,  F  and  F'  in  Fig.  111. 
That  diameter  A  A'  on  which  the  foci  are  located  is  called 
the  major  axis  and  is  taken  equal  p 

to  2  a,  i.e.  OA  =  a.  Similarly  the 
other  axis,  called  the  minor  axis, 
is  perpendicular  to  A  A'  at  0  and 
is  taken  equal  to  2  6,  i.e.  OB  =  h 
(Fig.  111). 

An  ellipse  may  be  constructed 
by  attaching  the  ends  of  a  string 
to  two  pins  fastened  in  a  board  or 
a  piece  of  stiff  cardboard  and  using  the  string  as  a  loop,  keep- 
ing it  stretched  by  means  of  a  pencil  and  allowing  the  pencil 
to  move.  Of  course  the  length  of  the  string  is  constant  — 
the  ends  being  fastened.  When  the  pencil  is  at  B,  Fig.  112, 
and  pins  at  F  and  F\  the  string  takes  the  position  FBF\ 
When  the  pencil  is  at  A',  the  string  reaches  from  F  to  A' 
and  back  to  F\  With  the  pencil  at  A,  we  deduce  AF  =  A'F'. 
In  either  case,  then,  the  length  of  the  string  is  equal  to  the 
length  of  the  major  axis,  2  a.     Thus  also  FB  =  F'B  =  a. 

The  perimeter  of  an  ellipse  is  given  approximately  by 
the  formula 

(12)  P=tr(a+b).* 

The  area  of  an  ellipse  is  given  exactly  by  the  formula 

(13)  A=Trab. 


*  A  more  exact  formula  for  the  perimeter  of  an  ellipse  is 

L        4:\a  +  bl    ^Q4.\a  +  bl  ^      J 

The  formula  x(a  -\-  b)  is  very  nearly  correct  if  o  —  6  is  small ;    i.e.  if  the 
ellipse  is  nearly  round. 


192  MATHEMATICS  [XI,  §  163 

Example,  Construct  an  ellipse  by  attaching  the  ends  of  a  string  4' 
long  to  two  pins  3^"  apart  fastened  in  a  cardboard.  By  measurement, 
the  major  axis,  2  a  =  4",  FB  =  OA  =  2",  and  OF  =  l\".     Thus 

OB  =  V22  -  (11)2  =  .97  =  1" 
approximately.     Hence  a  =  2",  h  =  I",  and  we  have 

Area  =  wab  =  (^)(2)(1)  =  6.3  sq.  in. 
Perimeter  =  7r(a  +  b)  =   (^)3  =  9.4". 

Many  things  are  elliptic  in  shape  or  in  cross-section.  For 
example,  sections  of  cupola  ventilators  placed  on  the  roof 
of  a  barn  are  ellipses.  Many 
photograph  mountings  and  in- 
terior decorations  are  elliptic. 
Many  arches  in  buildings, 
bridges,  etc.,  are  elliptic.  Heat- 
conducting  pipes  for  furnaces 
have  many  sections  that  are 
elhptic.  Fig.  112 

The  paths  or  orbits  of  the  earth  and  the  other  planets 
about  the  sun  are  rather  round  ellipses  with  the  sun  at  one 
focus.  Hence  the  earth  is  nearer  the  sun  during  a  part  of 
the  year  and  it  moves  faster  then  because  of  the  increased 
attraction.  When  is  the  earth  nearest  the  sun?  How 
much  nearer  is  it  at  that  time  ? 

EXERCISES 

1.  Find  the  area  and  the  perimeter  of  an  elliptic  photograph  6" 
long  and  4"  wide. 

2.  A  carpenter  wishes  to  construct  a  frame  for  supporting  the  brick- 
work while  laying  up  an  arch  over  a  grate,  the  arch  to  be  half  of  an 
ellipse  46"  wide  with  a  rise  of  11".  Draw  the  ellipse  to  scale  and  show 
how  to  locate  the  foci.     Find  their  distance  from  the  center  of  the  ellipse. 

3.  Compute  the  average  velocity  of  the  earth  per  second  in  its  orbit 
about  the  sun,  having  given  that  its  nearest  and  farthest  distances  from 
the  sun  are  90  millions  and  93  millions  of  miles  respectively. 


XI,  §  164]         GEOMETRY  —  MENSURATION 


193 


4.  Elliptic  gears  will  mesh  with  each  other  when  revolving  about 
their  foci. 

At  what  distance  from  the  end  of  the  major  axis  should  the  hole 
for  the  center  of  revolution  be  drilled  in  an  elliptic  gear  whose  axes 
are  8"  and  10"? 

Give  an  account  of  the  elliptic  gears  used  on  a  haypress  to  operate 
the  plunger. 

5.  How  large  an  opening  —  in  area  and  perimeter  —  is  necessary 
to  install  a  circular  ventilating  cupola  36"  in  diameter  on  a  roof  which 
has  a  pitch  of  5  in  24  ? 

How  many  7"  circular  intakes  will  be  necessary  to  equal  approxi- 
mately the  area  of  the  opening  in  the  roof? 

164.  Truncated  Prism.  A  truncated  prism  is  the  part 
of  a  prism  included  between  the  base  and  a  section  made  by 
a  plane  oblique  to  the  base. 

The  volume  of  a  right  truncated  triangular 
prism,  Fig.  113,  is  equal  to  the  product  of  one 
third  the  sum  of  its  lateral  edges  by  the  area 
of  the  base  to  which  those  edges  are  perpendic- 
ular, i.e. 
(14)  Volume=\{a-{-b+c)Xbase. 

The  same  formula  holds  for  any  triangular 
truncated  prism  (Fig.  114)  if  the  base  be  replaced 
by  the  area  of  a  section  perpendicular  to  the  edges. 

Example.  Corn  is  piled  into  the  corner  and  along  the 
side  of  a  square  building.  It  measures  4'  high  and  6'  X 
21'  on  the  floor  and  15'  along  the  top  crest.  Find  the 
volume  of  the  pile. 

The  pile  takes  the  shape  of  a  right  truncated  triangular 
prism.     Accordingly  its  volume  is 

1(15+21+21) .  i  (4x6)  =228  cu.  ft. 


^7 


N 


M 
Fig.  114 


Fig.  115 


194 


MATHEMATICS 


\X1,  §  164 


EXERCISES 

1.  In  excavating  on  the  side  of  a  hill  for  a  building  the  following 
measurements  were  taken.  From  the  surface  to  the  bottom  level  of 
the  cut  12',  9',  8',  6',  respectively,  at  the  four 

corners  of  the  rectangular  bottom  which  is  to 
be  30'  X  40'.  Find  the  amount  of  dirt  to  be 
removed. 

[Hint.  The  shape  of  the  required  exca- 
vation will  be  that  of  two  truncated  right 
triangular  prisms.] 

2.  What  is  the  volume  of  the  part  of  a  rectangular  barn  under  the 
roof  and  above  the  eaves  if  the  barn  measures  30'  X  60'  and  the  hip  roof 
has  a  pitch  of  1  in  3  on  sides  and  ends  ? 

Find  also  the  angle  made  by  the  roof  with  the  horizontal. 


Fig.  116 


Fig.  117.  —  Vertical  Sections  of  a  Ditch 

3.  Find  the  number  of  cubic  yards  of  gravel  in  a  pile  8'  wide  and 
10'  long  if  the  sides  and  ends  have  a  slope  of  2  in  5. 

4.  A  cut  for  a  ditch  has  the  dimensions  shown  in  Fig.  117,  which 
shows  the  longitudinal  vertical  section  and  the  vertical  cross-sections 
at  each  end.     Find  the  volume  of  the  earth  to  be  removed. 


165.  The  Sphere.  Definitions.  A  section  of  the  sphere 
made  by  a  plane  through  the  center  of 
the  sphere  is  called  a  great  circle;  if  the 
plane  does  not  pass  through  the  center 
of  the  sphere  the  section  so  made  is  a 
small  circle.  Parallels  of  latitude  on 
the  surface  of  the  earth  are  small  circles. 
The  meridians,  which  pass  through  both 
north  and  south  poles,  are  great  circles. 


XI,  §  166]        GEOMETRY  — MENSURATION 


195 


A  segment  of  the  sphere  is  a  portion  of  the  volume  of  a 
sphere  included  between  two  parallel  planes.  It  thus  has 
two  circular  bases.  In  case  one  plane  is  tangent  to  the  sphere, 
one  base  reduces  to  a  point.  The  segment  is  then  called  a 
segment  of  one  base.  In  Fig.  118,  the  part  of  the  sphere 
above  the  plane  DEF  is  a  segment  of  one  base;  that  be- 
tween DEF  and  ABC  is  a  segment  of  two  bases. 

A  zone  is  a  portion  of  the  surface  of  a  sphere  included 
between  two  parallel  planes  (Figs.  118,  119). 

The  altitude  of  the  segment  or  of  the  zone  is  the  perpen- 
dicular distance  between  the  planes. 


166.  Volume  of  a  Spherical  Segment.    Area  of  a  Zone. 

(15)  Volume  =  l'Trh{Z  ri^-fS  rzH/i'), 
where  h  denotes  the  altitude  of  the  segment, 
and  Tiy  Ti  denote  the  radii  of  the  bases, 
respectively. 

(16)  Lateral  Area  of  Zone  =  2Trrhf 
where  r  denotes  the  radius  of  the  sphere. 


FiQ.  119 


Example.     A  sphere  8"  in  radius  is  cut  by  two 
parallel  planes,  one  passing  2"  from  the  center,  the       ^ 
other  6"  from  the  center.     Find  the  volume  of  the    ^ 
segment  and  the  area  of  the  zone  between  the  two 
planes. 

The  lateral  area  of  the  zone  is 

2  7rr/i  =  2  7r(8)(4)  =64  7r=  201  sq.  in. 


Fig.  120 


To  find  the  volume,  the  radii  n,  r2  must  be  found.     From  the  ac- 
companying figure 

r2=EF  =  V^pI^^  =  V^^^  =  V28. 
Volume  =  i  nhiS  n^+S  ri'+h^)  =|  7r4(180+84  +  16)  =586  cu.  in. 


196 


MATHEMATICS 


[XI,  §  166 


EXERCISES 

1.  The  diameter  of  a  sphere  is  6  feet.  Find  the  surface  of  a  segment 
whose  height  is  2  feet. 

2.  The  radius  of  a  sphere  is  3  feet.  Find  the  height  of  a  zone  whose 
area  is  equal  to  that  of  a  great  circle. 

3.  Find  the  area  of  a  zone  of  one  base  whose  radius  is  2  feet  and  the 
height  of  the  zone  is  1  foot. 

4.  Find  the  volume  of  an  apple  whose  circumference  is  12  in. 

5.  Find  the  volume  and  the  surface  of  a  spherical  segment,  if  the 
radii  of  its  bases  are  6'  and  8',  respectively,  and  its  height  is  3'. 

6.  How  many  pints  of  water  can  be  put  in  a  wash-basin  in  the  shape 
of  a  segment  of  a  sphere,  if  the  distance  across  the  top  is  12  inches  and 
the  greatest  depth  is  4  inches? 

7.  What  is  the  volume  and  the  entire  surface  of  a  dishpan  in  the 
form  of  a  segment  of  a  sphere,  if  the  distances  across  the  top  and  bottom 
are,  respectively,  20"  and  16",  and  the  depth  is  5^"? 

8.  Compute  the  capacity  of  the  milk-can  shown  in  the 
diagram  if  it  has  the  following  dimensions,  expressed  in 
inches,  D  =  13",  d  =  r',  a  =  \",  6  =  4",  h  =  l^". 

9.  How  far  from  a  sphere  of  radius  6"  must  a  light  be 
placed  so  that  one  sixth  of  the  surface  of  the  sphere  will 
be  illuminated  ?  Fig.  121 


jS 


167.    Ellipsoid.     The  ellipsoid  may  be 
defined  as  the  solid  whose  section  by  any 
plane  is  an  ellipse,  Fig.  122. 
(17)  Volume  =  4/3  irabc, 

where  a,  6,  c  are  the  semi-axes. 


Fig.  122.  —  Ellipsoid 


168.   Paraboloid  of  Revolution.     This  is  a  solid  generated 
by  revolving  a  portion  of  a  parabola  about  its 
axis.  Fig.  123. 

(18)  Volume  =  T:fh/2, 

where  h  is  the  height  of  the  paraboloid  and 
r  is  the  radius  of  the  base. 

The  surface  of  such  a  paraboloid  is  the   ^^^  ^23  —  par- 
usual  reflecting  surface  for  headlights,  etc.  aboloid 


XI,  §  170]        GEOMETRY  —  MENSURATION 


197 


169.  Anchor  Ring  or  Torus.  A  torus  or  ordinary  anchor 
ring  is  generated  by  a  circle  of  radius  r  revolving  about  an 
axis  in  its  plane  (Fig.  124).  If  c  is  the  distance  of  the  center 
of  the  circle  from  the  axis  of  rotation, 
and  r  is  the  radius  of  the  circle,  the  vol- 
ume and  the  surface  of  the  anchor  ring 
are  given  by  the  formulas 

(19)  FoZume=2ir2cr2, 

(20)  Sur/ace  =  4Tr2cr. 


Fia.  124.  —  Anchor 
Ring 


B 

Fig.  125.  — a 
Prismoid 


170.  The  Prismoid  Formula.  A  solid  is  called  a  prismoid 
if  it  has  for  bases  any  two  polygons  in  parallel  planes  and 
for  faces  quadrilaterals  or  triangles.  Fig.  125. 
The  altitude  of  a  prismoid  is  the  perpendicular 
distance  between  the  planes  of  its  bases.  The 
mid-section  of  a  prismoid  is  that  section  made 
by  a  plane  parallel  to  and  midway  between  its 
bases.  The  mid-section  bisects  the  altitude 
and  all  of  the  lateral  edges. 

The  volume  of  any  prismoid  is  given  by  the  following 
simple  formula 
/2i)  Volume=  {b-\-  B-\-^  M)- , 

where  h  denotes  the  altitude  of  the  prismoid,  b  and  B  denote 
the  area-s  of  the  two  bases,  and  M  denotes  the  area  of  the 
mid-section. 

If  the  area  of  one  base,  h,  is  zero,  that  base  consists  of  a 
single  point  or  a  line,  and  the  prismoid  resembles  a  pyramid, 
a  cone,  or  a  wedge.  For  such  a  solid,  M=J5/4,  and  the 
formula  becomes  i. 

If  the  bases,  h  and  B,  are  equal,  the  prismoid  resembles 
a  prism  or  a  cylinder,  and  the  formula  becomes  V=hXB. 


198  MATHEMATICS  \X.l,  §  170 

A  frustum  of  a  pyramid  or  of  a  cone  is  a  prismoid  for  which 

2 


i.e.  4:M  =  h-\-B-\-2VbB 

since  the  mid-section  of  a  frustum  is  similar  to  and  midway 
between  the  bases  *  ;  and  the  formula  becomes 

h 

The  prismoid  formula  apphes  to  the  volume  of  a  sphere, 
where  h=  B  =  0,  and  M  is  the  area  of  an  equatorial  section 
of  the  sphere. 

EXERCISES 

1.  Compute  the  volume  of  a  football  whose  shape  is  an  ellipsoid  of 
semi-axes  3^  X  3^  X  5^. 

2.  Compute  the  solid  contents  in  cubic  inches  of  an  egg  whose  larger 
end  is  a  semi-ellipsoid  whose  semi-axes  are  |",  ^|",  and  |f",  and  whose 
smaller  end  is  a  semi-eUipsoid  whose  semi-axes  are  V",  ^|",  and  {^". 

3.  Find  the  volume  of  a  reflecting  surface  whose  form  is  a  paraboloid, 
if  its  height  is  5"  and  the  radius  of  the  base  is  3^". 

4.  The  cross-section  of  a  sohd  wrought-iron  ring  is  a  circle  of  3" 
radius.  The  inner  radius  of  the  ring  is  18".  Find  the  surface  and  the 
volume  of  the  ring. 

5.  Find  the  surface  and  volume  of  a  life  preserver  whose  shape  is 
an  anchor  ring,  formed  by  revolving  a  circle  of  a  radius  3"  about  an 
axis  13"  from  the  center  of  the  circle. 

What  weight  of  water  would  be  displaced  by  such  a  preserver  if  it  is 
submerged  ? 


*  If  s  and  s'  are  corresponding  sides  of  b  and  B,  then  (s  +  s')/2  is  the  cor- 
responding side  of  M.     Therefore, 

^        -^       and  ^'  ^ 


(8+sO/2      v^  (s  +  s')/2      VM 

Adding  these  two  equations  and  reducing,  we  find 

2  _V^+VrB^ 
1  Vm 


XI,  §  170]        GEOMETRY  —  MENSURATION 


199 


6.  A  barrel  made  of  circular  staves  has  an  end  diameter  of  18" 
and  a  bung  diameter  of  22".  It  is  34"  high.  What  is  its  capacity  in 
bushels,  and  in  gallons? 

[Hint.    Use  (8),  §  31.] 


10'6" 


Fig.  126. 


10'6" 

Sections  of  a  Railroad  Cut 


10'6" 


7.  If  the  barrel  of  Ex.  6  is  standing  on  end,  and  if  it  is  partly  filled 
with  a  Hquid  to  a  depth  of  24",  find  the  amount  of  the  liquid  in  gallons. 

[Hint.  Calculate  the  amount  of  the 
contents  above  the  half-barrel  by  the 
prismoid  formula.] 

8.  A  railroad  cut  has  the  dimensions 
shown  in  Fig.  126,  which  shows  the  vertical 
section  and  three  cross-sections.  Find  the 
number  of  cubic  yards  of  earth  to  be  re- 
moved in  this  cut. 

9.  Find  the  number  of  cubic  yards  of 
concrete  in  a  pier  having  the  dimensions  as 
shown  in  Fig.  127,  the  bases  being  rec- 
tangles with  semicircles. 


A  Pier 


CHAPTER  XII 

SOLUTION    OF   OBLIQUE   TRIANGLES 

171.  Construction  of  Oblique  Triangles  with  Ruler  and 
Compass.  A  triangle  consists  of  six  parts  —  three  sides 
a,  by  c,  and  three  angles  A,  B,  C,  respectively,  opposite  these 
sides.     These  parts  are  related  as  follows. 

I.  A-fBH-C=180°. 

II.  The  sum  of  any  two  sides  is  always  greater  than  the 
third  side. 

III.  The  larger  side  is  opposite  the  larger  angle,  i.e.  if  a>6, 
then  A>B,  etc. 

If  any  three  parts  of  a  triangle  are  given,  including  always 
at  least  one  side,  the  triangle  may  be  constructed.  The 
following  cases  arise. 

Case  I.     Given  One  Side  and  Two  Angles. 

Let  c  he  the  given  side,  and  let  A  and  B  he  the  given  angles 
(Fig.  128).     Take  EF  =  c,  Fig.  128,  and  construct   Z  A  at 


Fig.  128 

E  and  Z  B  at  F,  on  the  same  side  of  c.  Produce  the  sides 
of  Z  E  Siud  Z  F  till  they  meet  at  C.  Then  EEC  is  the 
required  triangle. 

The  third  angle  may  be  found,  before  constructing  the 
triangle,  from  theorem  I.     If  this  angle  is  measured  with  a 

200 


XII,  §  171]  OBLIQUE  TRIANGLES  201 

protractor  after  the  construction  is  finished,  the  accuracy 
of  the  construction  may  be  judged. 

Case  II.    Given  Two  Sides  and  the  Included  Angle. 
Let  b  and  c  be  the  two  given  sides  and  Z  E  their  included 
angle  (Fig.  129). 


FiQ.  129 

Lay  off  AB  =  c.  Fig.  129.  At  A,  construct  Z  BAD=  Z  E, 
On  AD  take  AC  =  b  and  draw  CB.  Then  A  ABC  is  the 
required  triangle. 

Case  III.    Given  the  Three  Sides. 

Let  the  three  sides  be  a,  b,  c  (Fig.  130). 

Take  BC  equal  to  a,  Fig.  130.  From  B  as  a  center,  with 
a  radius  equal  to  c,  describe  an  arc.  From  C  as  a  center, 
with  a  radius  equal  to  6,  describe  an  arc  intersecting  the 


n4— 

6 

— 

bZ 

/              \ 

a 

Fig. 

130 

other  arc  at  A.     Draw  AC  and  AB.    Then  BAC  is  the 
required  triangle. 

Case  IV.  Ambiguous  Case.  Given  Two  Sides  and 
an  Angle  Opposite  One  of  Them. 

Let  a  and  b  be  the  given  sides  and  A  the  angle  opposite  the 
side  a  (Fig.  131). 

Different  constructions  are  possible  according  as  a<6, 
a  =  b,  a  =  the  altitude  on  c,  or  a<the  altitude  on  c. 


202  MATHEMATICS  [XII,  §  171 

//  a<h  and  ZA  acute,  there  may  be  two  constructions. 
Lay  off  a  line  AE,  Fig.  131.  At  A,  construct  the  given 
angle  A  and  lay  off  AC=b.    From  C  as  center,  with  a 


radius  equal  to  a,  describe  an  arc  intersecting  AE  at  B  and 
B'.  Draw  CB  and  CB'.  Both  triangles  ACB  and  ACB' 
fulfill  the  conditions ;  there  are  two  possible  constructions. 

If  a<h  and  Z  A  obtuse,  there  is  only  one  construction. 
The  constructions  for  a  =  b,  a<b,  a<the  altitude  on  c,  are 
left  as  exercises  for  the  student. 

This  case  is  known  as  the  ambiguous  case.     Why  ? 

EXERCISES 

1.  Given  angles  of  50",  75°  and  a  side  2"  long  in  a  triangle ;  construct 
the  triangle,  using  rule  and  protractor.  What  is  the  third  angle? 
Measure  the  other  two  sides. 

2.  Given  A  =40°,  C  =  110°,  and  c  =  2.5"  in  a  triangle;  construct 
the  triangle.     Find  B  and  measure  a  and  b. 

3.  Construct  triangles  to  the  scale  1  rod  =  ^'',  having  given  a  =  14 
rods,  6  =  10  rods,  B  =  30°. 

Construct,  if  possible,  triangles  having  given 

4.  a  =  14,  6  =  6,  B  =  30°.  5.   a  =  14,  6  =  16,  B  =  30°. 
6.   a  =  14,  6  =  6,  C  =  30°.                           7.   a  =  14,  6  =  6,  c  =  16. 

172.  Solution  of  Oblique  Triangles.  To  solve  a  triangle 
is  to  compute  the  unknown  parts  when  any  three  parts  are 
given,  at  least  one  of  which  must  be  a  side.  This  is  one  of 
the  chief  uses  of  trigonometry.  An  aid  to  such  a  computa- 
tion is  the  construction  of  the  given  data  to  scale  (§  171). 


XII,  §  173] 


OBLIQUE  TRIANGLES 


203 


The  different  cases  that  arise  are  those  of  §  171.  If  the 
given  data  are  not  sufficient,  for  example,  if  a=  10, 6  =  6,  c= 3, 
the  triangle  is  not  solvable.     (See  II,  §  171.) 

Oblique  triangles  can  always  be  solved  by  dividing  the 
triangle  into  two  right  triangles  by  drawing  an  altitude,  and 
solving  the  resulting  right  triangles  by  previous  methods. 
(See  §§37-41.)  However,  this  process  is  not  always  the 
most  direct.  The  relations  given  in  §§  173-174  will  aid 
greatly  in  the  solution  of  the  different  cases. 


173.   The  Law  of  Sines.     The  sides  of  a  triangle  are  pro- 
portional to  the  sines  of  the  opposite  angles,  i.e. 

(1)  _JL_  =  _^  =  _^. 

sin  A     sin  B     sin  C 

Proof.     Let  ABC  be  any  triangle.  Fig.  132,  and  let  p  be 
the  altitude  on  6.     Then  in  all  cases  we  have 


6         c 
Fig.  132  h 


C  b  A 

Fig.  132  c 


(2) 


sinA  =  ^,     and     sinC=-2- 
c  a 


(3) 


By  dividing,  we  have 
sin  A     a 


or 


sin  C      c '       '     sin  A     sin  C 

If  the  perpendicular  were  dropped  on  c,  the  same  argu- 
ment would  give  a/sin  A  =  6/sin  B.     This  with  (2)  gives 


sin  A      sin  B     sin  C' 
which  is  known  as  the  law  of  sines. 


204  MATHEMATICS  [XII,  §  174 

174.  The  Law  of  Cosines.     In  §  158,  we  saw  that 

(4)  a''  =  ¥+c'-2  bx,    if  Z  A  is  acute  (Fig.  133  a), 
and 

(5)  a2=62+c2+2  hx,    if  Z  A  is  obtuse  (Fig.  133  b). 


Fig.  133  a  Fig.  133  b 

But  x  =  c  COS  A,  if  A  is  acute  and  x=  —c  cos  ^  if  A  is 
obtuse.  Substituting  these  values  of  x  in  (4)  and  (5),  re- 
spectively, we  have,  for  all  values  of  A,  acute  or  obtuse, 

(6)  a2=  62+^2-2  be  cos  A, 
or 

(7)  cos  A 


b^+c^-a^ 


2bc 
Either  of  these  formulas  is  called  the  law  of  cosines.   Similarly, 

62  =  0^2+02-2  ac  cos  B,     or     cos  g=  ^'^^'~^' : 

2ac       ' 

and 

c2  =  a2+62_2a6cosC,     or    cos  C=  ^'"^^'~^'- 

2  ab 

176.  The  Use  of  Logarithms.  Logarithms  may  be  used 
to  shorten  computations  involving  multiplications,  divisions, 
raising  to  powers,  extraction  of  roots.  They  are  often  quite 
convenient  but  they  are  not  essential.  In  the  solution  of 
triangles,  the  trigonometric  functions  are  nearly  always 
involved ;  hence,  in  the  logarithmic  solution,  we  need  to 
find  the  logarithms  of  these  trigonometric  functions. 

For  example,  if  the  hypotenuse  c  of  a  right  triangle  and 
an  acute  angle  A  are  given,  then  the  side  a  is 

a=c  sin  A. 


XII,  §  176]  OBLIQUE  TRIANGLES  205 

To  use  logarithms  to  compute  a  we  should  then  find  log  c 
and  log  (sin  A).     Now  sin  A  is  a  number  and  log  (sin  A) 
means  the  logarithm  of  that  number.     Thus,  if  A  =  26°,  then 
sin  26°=  .4384,  and  log  (.4384)  =  9.6418-10.    Hence 
log  (sin  26°)  =  9.6418-10. 

However,  this  double  use  of  the  table  is  not  necessary,  since 
these  and  similar  results  for  other  trigonometric  functions 
are  already  tabulated  in  the  table  of  trigonometric  functions 
in  the  columns  headed  Log.  Thus,  in  Table  IX  opposite 
26°,  under  Sine,  Log,  we  find  .6418.  The  characteristic 
9 (-10)  must  be  supplied  as  in  the  case  of  the  logarithm  of 
any  number,  by  noticing  the  position  of  the  decimal  point 
in  the  value  of  the  sine. 

Examples.     By  use  of  the  table,  verify  that 

log  sin  15°  =  9.4130  log  tan  76°  =  0.6032 

log  cos  26°  =  9.9537  log  tan  86°  =  1.1554 

log  tan  26°  =  9.6882  log  sin    4°  =  8.8436 

In  these  examples,  as  is  usual  in  computations,  the  final  -10  is  omitted. 

176.  Checking  the  Solution.  When  a  solution  to  a 
problem  has  been  obtained,  some  test  of  its  correctness 
should  be  applied.  In  general  this  may  be  done  by  obtaining 
the  same  solution  by  a  different  method  and  by  examining 
the  reasonableness  of  the  solution. 

In  checking  oblique  triangles,  the  parts  must  be  related 
as  in  I,  II,  III,  §  171.  A  very  great  aid  in  testing  a  solution 
is  the  construction  of  the  data  to  scale. 

In  applying  logarithms  to  the  checks,  the  logarithm  of  the 
computed  part  should  be  looked  up  in  the  table  instead  of 
using  the  logarithm  from  which  the  result  was  obtained. 
Logarithmic  checks  are  not  to  be  held  to  a  closer  agreement 
than  the  data  call  for.  In  solving  triangles,  except  when 
three  sides  are  given,  there  are  three  logarithms  to  be  added 


206  MATHEMATICS  \X.ll,  §  176 

(subtracted).  Hence  the  last  place  in  the  sum  of  these  log- 
arithms is  in  doubt  by  1.5  (§§72,  73).  When  three  sides 
are  given,  four  logarithms  may  be  combined.  It  is  usually 
sufficient  if  the  logarithmic  checks  agree  within  2  in  the 
last  place. 

Example.     In  a  right  triangle  with  the  following  data,  find  the 
unknown  parts  and  check  the  results. 

Given  the  hypotenuse  c  =  27.3,  Z  5  =  35°  15',  to 
find  Z  A,  b,  c. 


Formulas : 

^r 

(a)  A=90°-B  =  54°45'. 

.  (6)    h  =  csmB. 

B               a 

(c)    a=ccosB. 

Fig.  134 

Solution  : 

Using  (6), 

Using  (c), 

log  c  =  1.4362 

log  c  =  1.4362 

log  sin  5  =  9.7613 

log  cos  5  =  9.9120 

log  6  =  1.1975 

log  a  =  1.3482 

6  =  15.8 

a  =  22.3 

Check  :   a  sin  B  =  6  sin  ^. 

log  a  =  1.3483 

log  6  =  1.1987 

log  sin  B  =  9.7613 

log  sin  A  =9.9120 

1.1096 

1.1107  ' 

Discussion.  The  given  data  are  three-figured ;  hence  the  computed 
results  are  taken  to  the  nearest  three-figure  approximations.  There- 
fore in  applying  the  check,  we  use  the  logarithms  of  15.8  and  22.3  as 
found  from  the  table,  and  not  the  logarithms  of  6  and  a  from  the  solu- 
tion. The  logarithms  in  the  check  agree  within  2  in  the  third  place  of 
mantissas,  which  is  all  that  is  required  for  three-figure  data. 

177.  Case  I.  Given  One  Side  and  Two  Angles.  The 
third  angle  is  easily  found  from  the  fact  that  A-\-B-\-C=  180°. 
Each  unknown  side  is  found  by  the  law  of  sines.  For  the 
sine  of  an  obtuse  angle,  use  the  sine  of  its  supplement,  §  §  92, 
93.  Check  by  the  law  of  sines,  using  in  it  that  part  not 
already  used  in  the  solution. 


XII,  §  177] 


OBLIQUE  TRIANGLES 


207 


Example.     Given  a  =  26.34,  A  =  80°  15',  C = 44°  27',  to  find  6,  c,  and  B. 
Formulas  : 

(a)  B  =  180°-(A+C)=55°18'. 


(6)    b 


g  sin  B 
sin  A 

o  sin  C 


(See  (1),  §  173.) 


(c)    c=     . 

sin  A 

Solution  : 
From  (6), 

log  a  =  1.4206 

log  sing  =  9.9149 

sum  =  1.3355 

log  sin  A  =9.9937 

log  6  =  1.3418 

6  =  21.97 

Check  :  6  sin  C  =  c  sin  B. 

log  21.97  =  1.3417 

log  sin  C  =  9.8453 

1.1870 


(See  (1),  §  173.) 


Fig.  135 
From  (c), 

log  a  =  1.4206 

log  sin  C  =  9.8453 

sum  =  1.2659 

log  sin  A  =9.9937 


log  c  =  1.2722 
c  =  18.72 


(See  (1),  §  173.) 


log  18.72  =  1.2723 
log  sin  B  =  9.9149 


1.1872 

The  solution  is  thus  valid  as  the  logarithms  here  differ  by  only  2  in 
the  last  place  of  decimals. 

EXERCISES 

Solve  and  check  each  of  the  following  oblique  triangles,  and  find 
the  area. 

1.  a  =  34.62,  A  =62°  30',  C  =  34°  10'. 

Ans.    6  =  38.77,  c  =  21 .92,  Area  =  376.9 

2.  6  =  62.32,  A  =37°  37',  B  =  42°  15'.        Ans.     a  =  56.57,  c  =  91.24 

3.  c  =  123.4,  5  =  29°  18',  C  =  134°  10'.    Ans.    a  =  48.96,  6=84.19 

4.  One  side  of  a  triangular  field  is  1234  feet  long.  The  adjacent 
angles  are  43°  43'  and  34°  34'.  Find  the  perimeter  of  the  field  and  its 
area  in  acres.  Ans.  Perimeter  =  2820  ft.  Area  =  7  acres  lacking  27 
sq.  ft. 

5.  A  balloon  is  directly  over  a  straight  road.  From  two  points 
2^  miles  apart  on  this  road,  its  elevation  is  observed  to  be  62°  and  25°, 
respectively.     Find  the  height  of  the  balloon.     There  are  two  cases. 

Ans.  Height  =  8184  feet  when  both  points  are  on  the  same  side  o| 
the  balloon. 


208 


MATHEMATICS 


P:il,  §  178 


178.   Case  II.     Given  Two  Sides  and  the  Included  Angle. 

The  triangle  can  be  divided  into  two  right  triangles,  one  of 
which  contains  two  of  the  given  parts,  by  drawing  the 
altitude  on  one  of  the  known  sides. 

Again,  if  the  data  are  not  complex,  perhaps  three-figured 
or  less,  the  third  side  of  the  triangle  may  be  found  directly 
by  use  of  the  law  of  cosines.  The  other  two  angles  may  be 
found  by  the  law  of  sines  and  the  part  of  this  formula  not 
already  used  may  be  used  as  a  check. 

Both  of  these  methods  will  be  illustrated  by  examples. 

Example  1.  Two  sides  of  a  triangle  are  16  and  22  and  their  included 
angle  is  53°  8'.     Find  the  third  side,  the  other  two  angles,  and  the  area. 

Given  6  =  16,  c  =  22,  A  =  53°  8',  to  find  o,  B,  and  C,  and  the  area 
of  ABC. 

Formulas  : 

(a)         a  =  V62+c2-2  6ccosA.     (§  174.) 
b  sin  A 


(c)    Area  = 


(§  173.) 


base  X  altitude 


A        6  =  16       o 
Fig.  136 


Solution  :   Using  the  law  of  cosines,  we  have 

a  =  V256+484-2(16)(22)(.6000)  =  V3l7;6  =  17.8 
From  (&),  we  find 


hence 


sin  B  =  M:i«  =.7191; 

17.8 

5  =  45°  59'. 


Likewise 

.    ^     22(.8000) 
sm  C  =  — ^  ^  ^     •  = 
17.8 

hence  C  =  81°  25'. 

Finally,  by  (c),  we  have 

Area  =  |  base  X  c  sin  A  =  ^16  X  22^.8000)  =  140.8 


XII,  §  178] 


OBLIQUE  TRIANGLES 


209 


Example  2.     Given  b  = 

41.02,  c 

=  45.49,  A  =62°  10 

',  to  find  a,  B, 

andC. 

Formulas  :   Having  drawn  the  altitude  as  shown  in 

B 

Fig.  137,  we  have 

(a)      x  =  cco8A. 

(b) 

h=c  sin  A. 

(c)       y  =  b-x. 

id) 
if) 

tsinC  =  h/y. 
a  =  /i-rsin  C 

/^  j    \ 

(c)      B  =  lSO-iA+C). 

A^:^h^.^yJ^ 

Solution  : 

Fig.  137 

From  (a),  we  have 

From  id), 

log  c  =  1.6580 

\ogh  = 

1.6046 

log  cos  A  =9.6692 

log  2/  = 

=  1.2963 

log  a;  =  1.3272 

log  tan  C  = 

=   .3083 

X  =  21.24 

C  = 

=  63°  49' 

whence 

whence 

2/ =  19.78 

B  = 

=54°  r. 

From  (6), 

From  CO, 

log  c  =  1.6580 

\ogh  = 

=  1.6046 

log  sin  A  =9.9466 

log  sin  C  = 

=  9.9529 

log /i  =  1.6046 

loga  = 

=  1.6517 

;i  =  40.24 

a  = 

=  44.84 

Check  :  6  sin  A  =  o  sin  B. 

log  6  =  1.6130 

loga  = 

=  1.6517 

log  sin  A  =9.9466 

log  sin  B  = 

=  9.9081 

1.5596 

1.5598 

EXERCISES 
In  each  of  the  following  exercises  find  the  remaining  parts,  check 
the  results,  and  find  the  area. 

1.  a  =  56.70,  6  =  45.20,  C  =  47°  45'. 

Ans.    c  =  42.56,  A  =80°  26',  B  =  51°  49',  Area  =  948.7 

2.  6  =  576.2,  c  =  431.6,  A=73°  18'. 

3.  a  =  318.4,c  =  94.20,  5  =  37°  37'. 

4.  c  =  14.61,a  =  6.234,  5  =  111°  11'. 

6.  Two  forces  of  18.2  lb.  and  22.4  lb.  act  at  a  point  on  a  body  and 
at  an  angle  of  57°  15'  with  each  other.  Construct  the  resulting  par- 
allelogram of  forces  and  find  each  of  the  diagonals  of  this  parallelogram. 

6.  B  is  40  miles  from  A  in  direction  N.  68°  W.,  while  C  bears  N.  17°  E., 
55.2  miles.     What  is  the  distance  and  direction  of  C  from  B  ? 

Ans.    65.3  miles,  N.  54°  39'  E. 


210  MATHEMATICS  [XII,  §  179 

179.  Case  III.  Three  Sides  Given.  Find  the  angles 
by  the  law  of  cosines  and  check  by  the  law  of  sines.  Any 
small  discrepancy  between  180°  and  A-\-B-\-C  should  be 
distributed  among  the  angles  so  as  to  satisfy  this  require- 
ment. 

Whenever  the  square  of  one  side  is  greater  than  the  sum 
of  the  squares  of  the  other  two,  thus  making  the  numerator 
in  (7),  §  174,  negative,  ignore  the  minus  sign  while  dividing, 
then  use  for  the  result  the  supplement  of  the  angle  found  from 
the  table. 

For  example,  if  a  =  19,  h=  13,  c=  10,  we  have 


cos  A  = 


2  be 


^169+100-361  ^,(3^33) 
260 

This  means  that  A  is  greater  than  90°,  and  A  is  surely  less 
than  180°.  Therefore  let  us  set  A' =  180°- A  and  look  up 
cos  A' =  +  .3538  We  find  from  the  tables  A' =  76°  57'; 
whence  it  follows  that 

A  =  180°- A' =  103°  3'. 

Example.    Given  a =42.1,  6  =  54.3,  c  =  50.7,  to  find  A,  B,  C,  and 
the  area. 
Formulas  : 

(c)  C  =  180°-(A+B). 

(d)  Area  =  ^  6c  sin  A. 
Solution  :   From  (a),  we  have 

,,3  ^  ^  (54.3)3  +  (50.7)^-(42.1)^  ^  ggO^ 
2(54.3)  (50.7) 

A  =47°  7'. 


XII,  §  179]  OBLIQUE  TRIANGLES  211 

From  (6), 

r>_a^-\-c^-b^ 

cos  £>= 

2ac 
=  .3266 

B  =  70°56'. 
C  =  180°-(A+5) 
=  61°  57'. 
Check  :  a  sin  C =c  sin  A. 

log  o  =  1 .6243  log  c  =  1 .7050 

log  sin  C  =  9.9457  log  sin  A  =  9.8649 

1.5700  1.5699 

The  check  agrees  in  the  third  place  of  mantissas.  This  is  sufficient, 
since  the  values  of  the  lengths  of  the  sides  a,  6,  and  c  were  given  only  to 
three  significant  figures. 

Finally,  from  (d),  the  area  of  the  given  triangle  may  be  calculated 
as  follows. 

log  6  =  1.7348 

log  c  =  1.7050 

log  sin  A  =9.8649 

3.3047 

log  2  =0.3010 

log  Area  =  3.0037 

Area  =  1009. 

EXERCISES 

In  each  of  the  following  exercises  determine  the  remaining  parts, 
check  the  results,  and  find  the  area. 

1.  a  =  14,  6  =  16,  c  =  10. 

Ans.    A  =60^  B  =  SV  47',  C=38°  13'. 

2.  a  =  122,  6  =  146,  c  =  112. 

3.  a  =  32,  6  =  47,  c  =  62. 

[Hint.     From  a  figure,  it  is  seen  that  C  is  obtuse.] 

4.  a  =  132.4,  6  =  101.7,  c  =  84.50 

5.  A  farmer  has  a  four-sided,  irregular  field.  If  the  comers  are 
denoted  by  A,  B,  C,  D  taken  in  order  around  the  field  and  if  AB  =  17, 
BC  =  12,  CD  =  13,  DA  =  10,  and  BD  =  1Q,  find  the  area  and  the  angle 
at  A. 


212 


MATHEMATICS 


[XII,  §  180 


Fig.  139 


180.  The  Law  of  Segments.  In  §  179,  the  triangle  with 
three  sides  given  was  solved  by  use  of  the  law  of  cosines. 
A  second  method  for  this  case, 
better  adapted  to  the  use  of 
logarithms,  will  now  be  given. 

Drop  a  perpendicular  from  A 
on  a,  dividing  a  into  two  seg- 
ments X  and  y,  as  shown  in 
Fig.  139.     From  the  resulting  right  triangles,  we  have 

(8)  h^—x^  =  c^—y'^,    or    x'^  —  y^^b^^c^, 
or 

(9)  (x+2/)(x-2/)  =  (6+c)(&-c). 

From  the  figure,  we  see  that 

(10)  x-\-y  =  a. 

Dividing  the  sides  of  equation  (9)  by  the  sides  of  (10), 
in  order,  we  obtain  the  relation 


(11) 


x-y 


(b+c)(b-c) 


Solving  (10)  and  (11)  by  addition  and  subtraction,  we  find 
(12) 


a  ■  {b-\-c){b-c) 
2  2a 


(13) 


y= 


(b+c)(b-c) 
2a 


But  we  know  that 


(14) 


COS  C  = 


b' 


and 


cos  B=^  ; 
c 


hence  B  and  C  can  be  found.     Then 
A  =  lSO°-{A-\-B). 
The  solution  may  be  checked  by  the  law  of  sines. 


XII,  §  180] 


OBLIQUE  TRIANGLES 


213 


Example.     Given  a  =  27,  6  =  24,  c  =  19,  to  find 
A,  B,  C. 
Formulas : 

(a)    x+y  =  a. 

(6)       ^-y^if>+C)(.b-C), 

a 
(Find  X  and  y  from  (a)  and  (6)  by  addition  and  subtraction.) 

(d)  cosB  =  y., 


cos  C  = 


Solution:  x-\-y  =27 

^_„=(4^=  7.963 


cos  5  = 


9.52 


19 

log  9.52  =  0.9786 

log  19  =  1.2788 

log  cos  B  =  9.6998 

3  =  59°  5Q' 

A  =76°  48'. 

Check  :   b  sin  C  =  c  sin  B. 
log  6  =  1.3802 
log  sin  C  =  9.8360 
1.2162 


17.48 
9.52 


cosC= 


17.48 


24 

log  17.48  =  1.2424 
log  24  =  1.3802 
log  cos  C  =  9.8622 
C  =  43°  16'. 


log  c  =  1.2788 

log  sin  B  =  9.9372 

1.2160 


EXERCISES 

Use  the  law  of  segments  to  solve  each  of  the  following  triangles. 
Check  your  answers. 

1.  a  =  42.6,  6  =  38.4,  c  =  27.2 

2.  0  =  39.42,6  =  47.82,  c  =  61.37 

3.  Use  the  law  of  segments  to  check  Exs.  3,  4,  §  177. 

4.  Under  what  angle  is  a  ship  850  feet  long  seen  by  a  man  who  is 
standing  on  a  wharf  which  is  600  feet  from  the  bow  of  the  ship  and  950 
feet  from  her  §tern? 


214 


MATHEMATICS 


\X.U,  §  181 


181.   Case  IV.    Given  Two  Sides  and  an  Angle  Opposite 
One  of  Them.     Let  the  given  parts  be 

a,  h,  A. 
Construction  of  the  given  data  to  scale  will  usually  indi- 
cate the  possibilities  of  this  case.     See  Case  IV,  §  171, 
c  c 


o  Solution 
One  Solution 
Two  Solutions     ^, 

Fig.  141.  —  Possible  Constructions  for  the  Ambiguous  Case 

//  a<b  and  Z  A  acute,  there  are,  Fig.  141, 

two  solutions  if  a>the  altitude  on  c,  i.e.  if  a>h  sin  A. 

one  solution  if  a  =  the  altitude  on  c,  i.e.  if  a  =  h  sin  A. 

no  solution  if  a<the  altitude  on  c,  i.e.  if  a<b  sin  A, 

If  a>h,  there  is  one  solution. 

If  Z  A  is  obtuse,  there  is  never  more  than  one  solution. 

A  second  angle,  Z  B  here,  is  always  ob- 
tained from  the  relation  a  sin  B  =  h  sin  A. 
As  found  from  the  table,  Z  B  is  acute. 
Then  B',  Fig.  141,  is  the  supplement  of  B 
since  A  B'CB  is  isosceles. 

This  is  often  called  the  ambiguous  case. 

Example.  Given  a  =  20,  c  =  25,  A  =  52°  40',  to 
find  h,  B,  C. 

The  construction  shows  that  two  triangles  ABC,  ABC  fulfill  the 
conditions. 

Formulas  : 


Fig.  142 


(a) 
(ft) 


sin  C  = 


h  = 


c  sin  A 

a 

asm  B 

sin  A 


(c) 
(d) 


h'  = 


180 -C. 

a  sin  B' 
sin  A 


XII,  §  181] 


OBLIQUE  TRIANGLES 


215 


Solution:  From  (a), 

log  c  =  1.3979 
log  sin  A  =9.9004 
sum  =  1.2983 
log  a  =  1.3010 
log  sin  C  =  9.9973 
C  =  83°40'. 
B  =  43°40'. 
From  (6), 

log  0  =  1.3010 
log  sin  B  =  9.8391 
sum  =  1.1401 
log  sin  A  =9.9004 
log  6  =  1.2397 
6  =  17.4 
Check  :  6  sin  C =c  sin  B. 
log  17.4  =  1.2405 
log  sin  C  =  9.9973 
1.2378 
Check  :  6'  sin  C  =c  sin  B'. 
log  13.0  =  1.1139 
log  sin  r  =  9.9973 
1.1112 


C  =  96°20'. 

(A+C)=31°. 


C'  =  180° 
B'  =  1S0° 
From  id), 

log  a  =  1.3010 

log  sin  B'  =  9.7118 

sum  =  1.0128 

log  sin  A  =9.9004 

log  6' =  1.1124 

6' =  13.0 

log  25  =  1.3979 

log  sin  B  =  9.8391 

1.2370 

log  25  =  1.3979 

log  sin  B^= 9.71 18 

1.1097 


EXERCISES 

In  each  of  the  following  exercises  find  the  remaining  parts  and  check 
the  results.  Determine  first  in  each  case  whether  there  is  one  solution, 
or  two  solutions,  or  none.     Find  the  area. 

1.  a  =  50.8,  6  =  65.9,  A  =41°  00'.  Ans.  B  =  58°  21',  C=80°  40', 
c  =  76.4,  B'  =  12r  40',  C'  =  17°  20',  c'=23.1.  Areas  1652,  499. 

2.  6  =  249.8,  c  =  409.1,B  =  37°  38'. 

3.  6  =  232,  0  =  107,5=26°  36'. 

4.  o  =  518.1,  c  =  694.7,  C  =  115°  42'. 

5.  6  =  385.7,  c  =  491.2,  B  =  46°  15'. 

6.  A  man  walking  along  a  road  due  east  sees  a  fort  10.2  miles  away 
and  bearing  N.  57°  E.  If  the  guns  of  the  fort  have  a  range  of  8  miles, 
how  far  may  he  go  (o)  before  he  comes  within  range  of  the  guns,  and 
(6)  before  he  is  again  out  of  range? 


216 


MATHEMATICS 


\XU,  §  181 


7.  A  50-ft.  chord  of  a  circle  subtends  an  angle  of  100°  at  the  center. 
In  the  larger  segment,  a  triangle  with  one  side  57.8  ft.  is  to  be  inscribed. 
Find  the  length  of  the  third  side  and  the  area  of  the  triangle.  How 
many  solutions  are  possible? 

182.   Summary. 


Case 

Given 

To  Find 

Formulas 

Check 

I. 

One  Side  and  Two 

C  = 

C  =  180°-(A-\-B). 

Angles 

a  = 

c  sin  A 

§  177 

as,  c  =  26.34, 
^=80°  15', 
fi=44°  17'. 

6  = 

sin  C 

,      c  sin  5 

6= • 

sin  C 

a  sin  B  =  b  sin  A. 

II. 

Two  Sides  and  the 

a  = 

IsT    Method:      for   simple 

Included  Angle 
as,  6  =  41.02, 

B  = 
C  = 

data. 

§178 

a=V62-|_cJ-2  6ccosA. 

c =45.49, 

r>      6  sin  A 
sin  B= 

A  =62°  10'. 

C=180°-(A+B). 

Cain  B  =  bamC. 

2d  Method  : 

y\ 

x=c  cos  A. 

e/.'V 

h=c  sin  A. 

y  =  b—x. 

/  X     \y\ 

tan  C  =  h/y. 

B  =  180-{A-\-O. 

A         b        C 

Fig.  143 

a-     ^    . 
sinC 

a  sin  B  =  b  sin  A. 

III. 

Three  Sides 

A  = 

IsT  Method:       for    simple 

as,  a  =  42.1. 

B  = 

data. 

§179 

6  =  54.3, 
c  =  50.7 

C  = 

cos^-^+^^-"'. 

2  6c 
cos5=etc. ;  cos  C= etc. 
2d    Method:    Law  of  Seg- 
ments. 

§180 

A 
6/V 

x-\-y  =  a. 

^..y-(b+cHb-c) 
a 
x  =  (add) 
y  =  (subtract) 

/.  lA 

C         a          B 

Fig.  144 

cosB=^;  cosC=?- 
c                    6 

^  =  180°-(B+C). 

IV. 

Two  Sides  and  an 

C  = 

.     ^      c  sin  A 
sm  C= • 

Angle  Opposite 

B  = 

§181 

One  of  Them 

as,  a  =  20, 
c  =  25, 
^=52°  40' 

6  = 

Thus  C  =  an  acute  angle, 
and    C'  =  180°-C. 

fl  =  180°-(A+O. 

,      a  sin  B 

and  known  as  the 

6  sin  C=c  sin  B. 

Ambiguous  Case 

sin  A 

XII,  §  182]  OBLIQUE  TRIANGLES  217 

MISCELLANEOUS   EXERCISES 

1.  P  and  Q  are  separated  by  a  hill.  To  find  QP  a  straight  Hne  AB 
is  run  through  Q  such  that  AQ  =756  ft.,  QB  =  562  ft.,  Z  QAP  =  47°  29', 
and  Z  QBP  =  57°  45'.     Find  QP. 

2.  A  man  observes  the  angle  of  elevation  of  a  hill  to  be  29°  18'. 
After  walking  one  mile  directly  toward  the  hill  on  a  level  plane,  he  finds 
that  the  angle  of  elevation  is  36°  31'.     Find  the  height  of  the  hill. 

3.  The  parallel  sides  of  a  trapezoid  are  36  and  52  and  the  non- 
parallel  sides  are  23  and  27.     Find  the  angles  and  the  diagonals. 

4.  A  surveyor  has  the  following  data.  AB  =  487  feet,  BC  =  179  feet, 
Z  CAB  =  2S°.     Show  that  something  is  wrong  with  the  notes. 

5.  If  a  billiard  table  is  a  feet  long  and  b  feet  wide,  at  what  angle, 
/,  must  a  ball  be  made  to  strike  the  long  cushion  in  order  that  after 
striking  each  cushion  in  turn  it  will  pass  through  its  original  position? 

Ans.     tan/=-- 
a 

6.  A  trough  is  10  ft.  long  and  its  cross-section  is  a  semi-circle  whose 
radius  is  1  ft.  Find  the  number  of  gallons  of  water  it  contains  when 
the  water  measures  4  in.  deep.     When  6  in.  deep.     When  9  in.  deep. 

7.  A  road  OA  is  9|  miles  long  and  makes  an  angle  of  31°  16'  with 
a  straight  beach  OX.  From  A  two  straight  roads  AB  and  AB'  each 
6  miles  long  run  to  the  beach.  Find  the  distance  along  the  beach  from 
O  to  the  nearer  of  the  points  B  and  B'.     (Harvard.) 

8.  What  angle  does  y  =  2x-\-6  make  with  the  x-axis?  What  angle 
does  3  y— 4  X  =  12  make  with  the  x-axis? 

9.  Plot  the  lines  2  x  +3  i/  =  12  and  4  x  -3  y  =  12.  Find  their  point 
of  intersection  and  the  angle  each  makes  with  the  x-axis. 

10.  In  the  accompanying  figure,  O  is  the 
center  of  revolution  of  the  crank  pin  of  a 
steam  engine,  I  is  the  connecting  rod,  P  is 
the  end  of  the  sUde  connected  with  the 
piston  head,  and  A  is  the  angle  of  revo-  "^  ^Fia.  145 
lution  POR.  What  is  OP,  when  ie  =  6", 
i  =  24",  and  A=30°?    AlsoifA=45°?    If  A  =  135°? 

Ans.  29";  27.9";  19.4" 


CHAPTER  XIII 
LAND   SURVEYING 

183.  Introduction.  To  survey  land  means  to  measure 
distances  and  angles,  to  calculate  from  these  measurements 
other  distances  and  angles,  to  locate  and  mark  the  corners, 
to  record  these  measurements  and  locations  so  that  a  plat 
or  map  can  be  made,  and  to  find  the  area  of  the  tract  or 
parcel  of  land. 

Sometimes  it  is  desirable  to  resurvey  an  original  tract. 
The  surveyor  then  seeks  to  find  the  original  corners,  to 
retrace  the  boundaries,  and  to  replace  any  landmarks,  such 
as  posts,  monuments,  etc.,  that  have  decayed  or  have  been 
destroyed.  In  addition  to  this  he  is  often  called  upon  to 
subdivide  tracts  of  land  into  smaller  parcels,  as  in  city  sub- 
divisions. 

184.  Instruments.  For  measuring  distances  surveyors 
usually  use  a  steel  tape  100  feet  long,  or  a  chain  which  is 
66  feet  =  4  rods  long  and  which  contains  100  links.  For 
these  and  other  units  of  measurement,  see  §  185.  The 
surveyor  also  uses  pins  to  mark  the  successive  applications 
of  the  tape  or  chain ;  range  poles  for  sighting  and  ranging 
out  directions ;  and  a  graduated  rod  or  pole  called  a  leveling- 
rod  provided  with  a  sliding  target  by  which  he  can  measure 
elevations  and  depressions,  as  in  establishing  the  grade  for 
a  ditch. 

Almost  all  of  the  old  land  surveys  of  this  country  were 
made  with  the  compass  as  the  angle-measuring  instrument. 

218 


XIII,  §  184] 


LAND  SURVEYING 


219 


However,  it  has  now  been  supplanted  to  a  great  extent  by 
the  transit. 

The  transit  (Fig.  146)  is  the  instrument  now  most  used  by 
surveyors  and  engineers  for  measuring  horizontal  angles, 
and,  with  certain  attachments,  for  measuring  vertical  angles 


Fig.  146 


and  distance,  and  for  leveling.     It  is  mounted  on  a  tripod 
and  consists  essentially  of  the  following  three  parts. 

(a)  the  telescope  to  determine  direction.  To  the  telescope 
is  attached  a  level  and  a  vertical  graduated  circle  to  determine 
angles  of  elevation  and  depression. 


220  MATHEMATICS  \XIU,  §  184 

(6)  the  alidade,  a  horizontal  circular  piece  provided  with 
vertical  supports  to  carry  the  telescope,  with  a  graduated 
circle,  and  with  precise  levels  so  that  this  circle  can  be 
brought  exactly  into  the  horizontal  plane.  Angles  in  this 
plane  may  be  read  on  this  circle  as  the  angular  turn  of  the 
telescope. 

(c)  the  magnetic  compass. 

186.  Surveyor's  Measures.  City  lots  are  usually  sur- 
veyed in  terms  of  feet  and  square  feet,  with  decimal  parts 
of  these  units.  However,  in  measuring  larger  pieces  of  land 
the  following  measures  are  used. 

1  rod  =  16i  feet. 

=  1  perch=  1  pole  (older  plan). 
1  chain  (ch.)        =4  rods  =  66  feet. 

=  100  links  ill). 
1  link  =7.92  inches. 

1  Acre  (A.)  =160  square  rods  (sq.  rd.). 

=  10  square  chains  (sq.  ch.). 
1  section  =  640  acres. 

In  some  states  Spanish-American  units  are  used  as  follows  : 

1  Vara   (va'-ra)    =  33^  inches  in  Texas, 

=  33  inches  in  California, 
=  32.9927  inches  in  Mexico. 

1  labor  (la-bor')  =a  square  whose  side  is  1000  varas, 
=  1,000,000  square  varas, 
=  173  acres  (approximately), 
=  a  quarter  section  (roughly). 

1  league  =  a  square  whose  side  is  5000  varas, 

=  25  labors, 
=  6.8  square  miles  (approximately), 

1  (Spanish)  -4 ere  =  5645  square  varas  (in  Texas), 


XIII,  §  186] 


LAND  SURVEYING 


221 


Since  100  links  =1  chain,  the  older  plan  of  writing  both 
chains  and  hnks  in  a  survey  is  discarded  and  distances  are 
written  in  terms  of  decimals  of  a  chain.  For  example,  11 
chains  24  links  is  written  11.24  ch.  Similarly,  the  area  is 
written  in  decimals  of  an  acre,  as  2  acres  4  square  chains  = 
2.4  A.,  2  A.  47  sq.  rd.  =  2.294  A. 

186.  General  Procedure  as  to  Comers,  Offsets,  Obstacles, 
Measuring  Slopes,  etc.  When  a  corner  has  been  determined 
in  a  survey,  and  agreed  upon  by  all  parties  concerned,  it 
should  be  marked  with  stones,  iron  or  concrete  posts,  etc., 
in  such  a  way  as  to  perpetuate  its  location  as  long  as  possible. 

In  measuring  distances  two  men,  a  leader  and  a  follow er^ 
carry  the  tape  or  chain.  The  leader  carries  ten  pins,  and 
at  each  measurement  he  "  sticks  "  a  pin  to  mark  it.  On 
moving  forward  the  follower  picks  up  this  pin.  The  process 
is  repeated  and  tallied,  both 
leader  and  follower  keeping 
their  own  tallies. 

Since  areas  are  computed 
as  if  the  land  were  flat,  it  is 
necessary  in  measuring  on  slopes  to  keep  the  chain  level 
(Fig.  147). 

It  sometimes  happens,  on  account  of  fences,  brush,  ponds, 
or  other  obstructions,  that  a  measurement  cannot  be  made 

c 

Ci \D 


Fig.  147.  —  Measuring  on  a  Sixjpe 


B        D 

Fig.  149 


on  the  desired  line.  The  surveyor  then  measures  a  con- 
venient portion  of  a  parallel  to  the  required  line  at  a  certain 
perpendicular  distance  from  it,  as  CD  =  BE^  Fig.  148. 


222  MATHEMATICS  fKIII,  §  186 

An  obstacle  sometimes  may  be  passed  and  the  general 
direction  continued  beyond  it  by  measuring  an  equilateral 
triangle  about  the  obstacle  as  in  Fig.  149. 

187.  Computing  the  Area.  In  §§18  and  42,  methods  for 
finding  the  area  of  irregular  shaped  tracts  have  been  discussed. 
These  methods  will  now  be  discussed  more  fully,  making 
use  of  the  latitudes  and  longitudes  of  the  corners  of  the  sur- 
vey. These  may  be  looked  upon  as  the  coordinates  of  the 
vertices  of  the  survey  and  follow  from  the  lengths  and  bear- 
ings of  the  courses. 

188.  Definitions.  If  the  y-axis  be  a  meridian  and  the 
X-axis  an  east-west  line,  then  the  ordinate  of  a  point  is  called 
its  latitude,  its  abscissa  is  its  longitude. 

Let  ABhe  any  course.  Fig.  150.  Draw  a  meridian  through 
A  and  perpendiculars  BB^,  ^ A'  to  it  and  the  y-Sixis.  Then 
PA  =  OA'  is  the  latitude  of  A,  OB'  is  the 
latitude  of  B,  and  thus  A'B'=the  latitude-  ci-^-¥- 
difference  of  the  course  AB.  Similarly  A  A', 
BB'  are  the  longitudes  of  A  and  B,  respec- 
tively; and  KB  =  the  longitude-difference  of 
the  course  AB. 

Latitude-difference  north,  i.e.  a  northing,  is 
taken  to  be  positive,  a  southing  is  taken  to  be 
negative.     Similarly,  longitude-difference  east,  i.e.  an  easting, 
is  positive,  a  westing  is  negative. 

The  double  longitude  of  a  course  is  the  sum  of  the  longi- 
tudes of  its  ends.  Thus,  in  Fig.  150,  the  double  longitude  of 
AB  is  AA'+BB'. 

189.  Area  when  Lengths  and  Bearings  of  Courses  are 
Given.  By  projecting  the  courses  of  a  tract  as  ABODE  A, 
Fig.  151,  on  a  meridian,  we  see  that  to  every  course  there 


XIII,  §  190] 


LAND   SURVEYING 


223 


Fig.  151 


corresponds  a  trapezoid.  Then  to  find  the  area  of  ABCDEA 
we  add  the  trapezoids  CC'DD'  and  DD'EE'  corresponding  to 
courses  which  hear  south  and  from  this  sum 
subtract  the  sum  of  the  trapezoids  correspond- 
ing to  courses  which  hear  north.  The  areas 
of  the  one  set  are  called  south  areas,  the 
others  north  areas. 

Again,  the  douhle  area  of  any  of  these  trap- 
ezoids as  AA'BB'  is  numerically  equal  to 
the  product  of  the  double  longitude  of  the  corresponding 
course  times  its  latitude-difference.  Thus  the  double  area 
of  AA'BB'={AA'+BB')A'B\  We  have,  therefore,  the 
following  rule. 

Rule  to  Find  the  Double  Area  of  a  Tract.  Multiply  the 
douhle  longitude  of  each  course  hy  its  latitude-difference,  plac- 
ing the  product  in  a  column  of  north  areas  when  the  hearing  is 
north,  and  in  a  column  of  south  areas  when  the  hearing  is  south. 
Add  up  the  two  columns  and  take  the  difference  of  the  results. 

190.  Latitude-differences,  Longitude-differences,  and 
Double  Longitudes.  From  Fig.  152,  KA  =  AB  cos  KAB; 
KB  =  AB  sin  KAB,  i.e.  for  any  course,  we  have 

Latitude-difference  =  length  of  the  course 
times  cosine  of  its  hearing, 

Longitude-difference  =  length  of  the  course 
times  sine  of  its  hearing. 

From  Fig.  152,  let  AB  and  BC  be  any 
two  successive  courses.  Then  the  double 
longitude  of  BC=  BB'+CC=(BB'+AA')  -^KB+MC, 

That  is,  the  double  longitude  of  any  course  equals  the 
douhle  longitude  of  the  preceding  course,  plus  the  longitude- 
difference  of  that  course,  plus  the  longitude-difference  of  the 
given  course,  eastings  counted  positive  and  westings  negative. 


Fig.  152 


224 


MATHEMATICS 


fXIII,  §  190 


If  the  reference  meridian  is  taken  one  chain  west  of  the 
most  westerly  corner  and  the  survey  begins  at  that  corner, 
then  for  the  first  course 

the  double  longitude  of  AB=  AA'+BB', 

•  =AA'^B'K+KB, 
=  1  + 1  +  longitude-difference  oi  AB. 

Thus,  double  longitude  of  the  first  course  equals  2  plus 
its  longitude-difference. 

//  the  reference  meridian  posses  through  the  most  westerly 
corner,  then  the  double  longitude  of  the  first  course  is  merely 
its  longitude-difference. 

191.  Balancing  the    Survey.     For   a   closed   survey   we 

should  have 

northings  =  southings, 

eastings  =  westings. 

If  these  are  not  equal,  the  difference  is  called  the  error  of 
closure.  This  error,  if  not  too  large,  may  be  distributed 
among  the  sides  in  proportion  to  their  respective  lengths  so 
that  the  survey  will  close.  This  is  called  balancing  the  survey. 
This  and  other  details  of  computation  and  its  arrangement 
will  be  illustrated  in  the  example,  §  192. 

192.  Illustrative  Example.  The  field  notes  for  a  tract  of 
land  are  given  in  the  following  table.  Plot  this  data  to  scale, 
balance  the  survey  for  error  of  closure,  and  find  the  area. 


Course 

AB 

BC 

CD 

DE 

EA 

Beabing 

N.  40°  E. 

N.  60°  E. 

S.  20°  E. 

S.  60°  W. 

N.  38i°  W. 

Distance 

7.18  ch. 

9.20  ch. 

15.22  ch. 

11.42  ch. 

12.72  ch. 

Plot  the  data  to  scale  and  take  the  reference  meridian 
one  chain  west  of  the  most  westerly  corner,  i.e.  AA'=1  ch. 
(Fig.  153). 


XIII,  §  192] 


LAND  SURVEYING 


225 


Latitude-differences  and  lon- 
gitude-differences must  next  be 
computed  and  balanced.  For 
the  first  course  AB,  we  have, 

lat.-diff.  =  (7.18)cos40° 

=  (7. 18)  (.7660)  =  5.50 

long.-diff.  =(7.18)  sin  40° 

=  (7.18)(.6428)=4.61 

These  and  the  results  for  the 
other  courses  may  be  tabulated  as  follows. 


FiQ.  153 


Ck)UR8E 

Beabimg 

Distance 
(Chains) 

Lat.- 

DiFF. 

Long. 

-DlFP. 

N. 

S. 

E. 

W. 

AB 

N.  40°    E. 

7.18 

5.50 

4.61 

BC 

N.  60°    E. 

9.20 

4.60 

7.97 

CD 

S.  20°    E. 

15.22 

14.30 

5.21 

DE 

S.  60°    W. 

11.42 

5.71 

9.89 

EA 

N.  38|°  W. 

12.72 

9.99 

7.87 

55.74 

20.09 

20.01 

17.79 

17.76 

20.01 

17.76 

Error 

.08 

.03 

The  error  in  latitude-differences  is  .08,  in  longitude-differ- 
ences .03.  The  survey  will  be  balanced  or  closed  by  distribut- 
ing these  errors  among  the  several  courses  in  proportion  to 
their  lengths.     The  perimeter  of  the  tract  is  55.74  ch. ;  hence 

.08 


error  in  latitudes  for  1  chain  = 


error  in  longitudes  for  1  chain  = 


55.74 

.03 
55.74 


.0014, 


.0005 


Therefore,  for  AB  =  7.1S  chains,  the  error  in  the  latitude 
is  assumed  to  be  (7.18)  (.0014)  =  .01,  and  the  error  in  the  lon- 
gitude is  assumed  to  be  (7.18)  (.0005)  =  .00 


226  MATHEMATICS  [XIII,  §  192 

Similar  results  may  be  computed  and  tabulated  for  the 
other  courses  BC,  CD,  DE,  and  EA. 

The  northings  are  too  large  and  each  must  be  diminished 
by  the  corresponding  corrections;  the  southings  are  too 
small  and  each  must  be  increased ;  similarly  for  the  eastings 
and  westings.  On  making  these  corrections  the  work  will 
appear  as  in  the  table  on  page  227. 

To  find  the  area,  double  longitudes  are  used.     These  are 

computed  as  follows  from  the  rule  in  §  190. 

long.-diff.  of  1st  course  =     4.61 

add  2. 

double  longitude  of  1st  course  =     6.61 

long.-diff.  of  1st  course  =     4.61 

long.-diff.  of  2d  course  =     7.97 

double  longitude  of  2d  course  =    19.19 

long.-diff.  of  2d  course  =     7.97 

long.-diff.  of  3d  course  =     5.20 

double  longitude  of  3d  course=   32.36 

long.-diff.  of  3d  course  =     5.20 

long.-diff.  of  4th  course  =  —9.90 

double  longitude  of  4th  course  =  27.66 

long.-diff.  of  4th  course  =  —9.90 

long.-diff.  of  5th  course  =  —7.88 

double  longitude  of  5th  course  =     9.88 

Check:  From  a  figure  it  is  easily  seen  that  the  double 
longitude  of  the  last  course  is  numerically  equal  to  its  longi- 
tude-difference plus  2.  This  furnishes  a  check  on  the  ac- 
curacy of  the  work.  If  the  reference  meridian  is  made  to  pass 
through  the  most  westerly  corner,  then  the  double  longitude  of 
the  last  course  is  equal  numerically  to  its  longitude-difference. 

These  values  of  the  double  longitudes  may  now  be  in- 
serted in  the  table  and  the  double  areas  computed  by  the 
rule  of  §  189.     The  work  then  takes  the  following  final  form. 


XIII,  §  192] 


LAND  SURVEYING 


227 


^ 

f 

(N  00 

CO  '"f 

88  S5 

1-H 

O 

1 

g 

< 

& 

CO  00 

il 

CO 

S2 

1— ( 

1 

05   i-H                    CO 

00 

(M      II 

.9 

^28        g 

c3 

s 

|i 

,-(  OS  CO  CO  00 

CO    rH    CO    CO   GO 

1 

*2?3S5* 

^' 

S§8 

S 

2 

•< 

pq 

OS  l^ 

I> 

W 

S  OS  (M 

'«1<  t>^  »o 

OQ 

(N  CO 
CO  t^ 

15 

^§      ^ 

§ 

lO  ^                  OS 

§ 

1 

^ 

OS  l^ 

00  00 

g 

OS  l> 

r-l 

C 

T-H   t^   ,-H 

CO  OS  (N 

S 

'^  t>  »o 

1-1 

a 

5 

i' 

CO  t^ 

1-H 

1-H 

o 

t  ^ 

lO  CO               OS 

§ 

•j; 

I?"" 

»0   TjH                     OS 

8 

g^ 

2S?3^S 

^ 

5^ 

i>  OS  u:)  i-H  (N 

s 

i 

tZ;  ^OQOJ  ^ 

1 

228 


MATHEMATICS 


PCIII,  §  192 


EXERCISES 

Plot  the  following  described  plats  of  land  to  scale,  balance  the  sur- 
vey for  error  of  closure,  and  find  the  area  in  acres. 
1. 


Course 

AB 

BC 

CD 

DE 

EA 

Bearing 

N.  21°  E. 

N.  65°  E. 

S.  33°  E. 

S.  65i°  W. 

N.  48°  W. 

Distance 

57.16  ch. 

7.68  ch. 

52.20  ch. 

50.96  ch. 

13.40  ch. 

Ans.     187.13  A. 


Course 

AB 

BC 

CD 

DE 

EA 

Bearing 

S.  23°  E. 

S.  80°  E. 

N.  38°  E. 

N.  42°  W. 

S.  60i°  W. 

Distance 

48  ch. 

52  ch. 

46  ch. 

62  ch. 

64  ch. 

Ans.     666.36  A. 


Course 

AB 

BC 

CD 

DE 

EA 

Bearing 

N.  41°  E. 

S.  70°  E. 

S.  29°  E. 

West 

N.  49i°  W. 

Distance 

60  ch. 

40  ch. 

72  ch. 

75  ch. 

48  ch. 

Ans.     32.89  A. 


4. 


Course 

AB 

BC 

CD 

DE 

EA 

Bearing 

N.  47°  E. 

East 

S.  32°  E. 

South 

N.  77°  53'  W. 

Distance 

964  ch. 

822  ch. 

644  ch. 

523  ch. 

1961  ch. 

Ans.     235.79  A. 


5.  In  the  example  of  §  192,  find  the  length  and  bearing  of  AC  and 
of  AD. 

6.  Assuming  that  the  measurements  in  the  following  description 
were  all  correctly  made,  find  the  length  of  the  last  course  and  compute 
the  area. 

Beginning  at  an  iron  pin  in  the  center  of  the  township  road  and  at 
the  northeast  corner  of  said  tract,  thence  N.  89°  W.,  140.8  rods  to  a 
stone  set  in  the  center  of  the  road,  thence  S.  29°  10'  W.,  65.7  rods  to  a 
large  boulder,  thence  S.  21°  18'  E.,  42.3  rods  to  an  8"  X  8"  concrete 
post  set  in  the  ground  on  the  south  bank  of  the  creek,  thence  S.  65° 


XIII,  §  194]  LAND  SURVEYING  229 

36'  E.,  38.6  rods  to  an  iron  pin  in  the  center  of  the  Salem-Brookville 
turnpike,  thence  N.  59°  14'  E.,  150.6  rods  to  an  iron  pin  at  the  cross- 
roads, thence  N.  2°  W.  to  the  place  of  beginning. 

7.  Balance  the  survey  and  compute  its  area  from  the  following 
field  notes, 

N.  27°  E.,  366.0  feet;  N.  19°  W.,  354.0  feet;  S.  69°  W.,  655.5  feet; 
S.  19°  E.,  275.3  feet ;  S.  71°  E.,  486.0  feet.  Ans.     6.732  A. 

193.  Locating  by  Metes  and  Bounds.  The  system  of 
describing  and  locating  lands  by  metes  and  bounds  is 
used  extensively  over  the  world  to-day  and,  quite  naturally, 
was  introduced  into  this  country  by  the  early  settlers  from 
Europe.  To  locate  a  desired  area  or  region  by  landmarks, 
some  permanent,  or  well-defined,  point  of  beginning  is  first 
established.  The  boundary  lines  are  then  described  in  direc- 
tion and  distance  with  reference  to  this  point  of  beginning 
and  with  reference  to  more  or  less  permanent  natural  objects 
such  as  trees,  streams,  well-established  highways,  as  well  as 
concrete  monuments,  stones,  iron  posts,  etc.,  placed  for  the 
purpose.  The  directions  are  by  magnetic  compass  and  the 
distances  are  in  terms  of  the  surveyor's  units.  However, 
the  fact  that  landmarks  decay  and  change  and,  that  the 
magnetic  direction  of  the  same  line  changes  from  year  to  yeary 
gives  a  transitory  character  to  the  platting  of  land  in  this 
way. 

194.  General  Plan  of  Government  Surveys.  In  1785 
practically  all  of  the  territory  north  and  west  of  the  Ohio 
River  had  been  ceded  to  the  United  States  Government  by 
the  withdrawal  of  state  claims.  A  law  passed  by  Congress 
May  20,  1785,  provided  that 

"  The  surveyors  .  .  .  shall  proceed  to  divide  the  said 
territory  into  townships  six  miles  square,  by  lines  running 
due  north  and  south  and  others  crossing  these  at  right  angles, 
as  near  as  may  be." 


230  MATHEMATICS  [XIII,  §  194 

Owing  to  the  convergence  of  the  meridians  thiy,  of  course, 
was  a  mathematical  impossibiUty.  The  phrase  "  as  near 
as  may  be,"  however,  has  been  broadly  interpreted.  Ac- 
cording to  the  provisions  of  this  act  and  the  acts  of  May  18, 
1796,  May  10,  1800,  and  Feb.  11,  1805,  and  to  the  rules  of 
the  commissioners  of  the  general  land  office,  a  complete 
system  exists,  whose  main  features  are  as  follows. 

First.  Land  is  to  be  surveyed  or  marked  out  in  divisions 
in  the  form  of  squares  or  rectangles.  This  has  been  done  in 
the  western  states,  including,  roughly,  all  of  the  states  of  the 
Union  except  the  original  thirteen  Colonies. 

Second.  .  To  mark  out  the  largest  squares,  north  and 
south  lines,  called  meridians,  are  first  laid  off  24  miles 
apart  and  marked  with  cornerstones,  trees,  or  other  per- 
manent objects.  East  and  west  lines,  called  base  lines, 
are  then  run  at  right  angles  to  these  meridians  at  intervals 
of  24  miles.  This  .would  divide  the  land  up  into  24-mile 
squares  were  it  not  for  the  convergence  of  the  meridians 
towards  the  poles  of  the  earth.  This  may  be  noticed  on 
maps  in  geographies  and  in  Fig.  154.  Each  24-mile  tract 
is  then  divided  into  sixteen  nearly  equal  squares  called 
townships  by  running  north-south  and  east-west  lines 
through  the  quarter  points  of  the  sides  of  the  large  tract. 

Certain  meridians  called  principal  meridians  are  located 
with  great  care.  They  govern  the  surveys  of  lands  lying 
along  them  for  considerable  distances  both  toward  the 
east  and  toward  the  west.  The  vertical  rows  of  townships 
parallel  to  the  principal  meridian  are  called  ranges.  The 
first  row  on  the  east  is  called  range  No.  1  and  is  written 
RIE ;  the  second  row  east  is  No.  2,  and  is  written  R2E,  etc. 
Similarly,  standard  base  lines  are  located.  The  tiers  of 
townships  parallel  to  this  base  line  are  numbered  north  and 
south  of  it.     A  township  in  the  first  tier  north  of  the  base 


XIII,  §  194] 


LAND  SURVEYING 


231 


line  is  township  No.  1  and  is  written  Tl  N ;  one  in  the  second 
tier  south  is  No.  2  and  is  written  T2Sj  etc. 

A  township  is  thus  identified  by  its  number  and  its  range 
from  some  standard  base  line  and  principal  meridian. 


Fig.  154.  —  Base  Line  and  Principal  Meridian 


TIN, 


Point  out  the  following  townships  in  Fig.   154. 
R2W;  TSN,  MW ;  T2S,  R7W ;  T3S,  RSE. 

Third.  Townships  are  required  to  be  divided  into  smaller 
squares  called  sections  one  mile  each  way  and  containing 
640  acres,  "  as  near  as  may  be.'*  These  sections  are  num- 
bered as  in  Fig.  155,  except  the  "  first  seven  ranges  "  laid 


6 

5 

4 

3 

2 

1 

36 

30 

24 

18 

12 

6 

31 

32 

33 

34 

35 

36 

7 

8 

9 

10 

11 

12 

35 

29 

23 

17 

11 

5 

30 

29 

28 

27 

26 

25 

18 

17 

16 

15 

14 

13 

34 

28 

22 

16 

10 

4 

19 

20 

21 

22 

23 

24 

19 

20 

21 

22 

23 

24 

33 

27 

21 

15 

9 

3 

18 

17 

16 

15 

14 

13 

30 

29 

28 

27 

26 

25 

32 

26 

20 

14 

8 

2 

7 

8 

9 

10 

11 

12 

31 

32 

33 

34 

35 

36 

31 

25 

19 

13 

7 

1 

6 

5 

4 

3 

2 

1 

Fig.  155 


Fig.  156 


Fig.  157 


off  by  Hutchins  west  of  the  Pennsylvania  line.  (See  §  196.) 
These  are  numbered  as  in  Fig.  156.  In  western  Canada 
the  sections  are  numbered  as  in  Fig.  157. 


232  MATHEMATICS  [XIII,  §  195 

195.  Standard  Parallels  or  Correction  Lines.  The  eastern 
and  western  boundaries  of  townships  are,  as  nearly  as  may- 
be, true  meridians.  Hence,  when  they  are  extended  north- 
ward through  several  tiers  of  townships  their  convergence 
becomes  considerable.  At  latitude  40°  the  convergence  is 
about  6.7  feet  per  mile  or  over  40  feet  to  each  township. 
To  prevent  this  diminution  in  the  size  of  townships  to  the 
north  of  the  base  line,  standard  parallels  are  run,  along  which 
six-mile  measurements  are  made  for  a  new  set  of  townships. 
These  lines  are  called  correction  lines  for  obvious  reasons. 

The  standard  parallel  or  correction  lines  have  been  run 
at  varying  distances,  the  present  distance  being  24  miles. 
Public  roads  are  usually  built  on  the  section  lines,  and 
wherever  a  north  and  south  road  crosses  a  correction  line 
there  is  a  "  jog  "  in  the  road  as  shown  on  Fig.  154. 

196.  Principal  Meridians  and  Base  Lines  of  the  United 
States.  The  first  public  land  survey  in  the  United  States 
was  started  in  1786  by  Thomas  Hut  chins,  Geographer  of 
the  United  States,  who,  with  thirteen  assistants,  laid  off  a 
line  from  the  southwest  corner  of  Pennsylvania  due  north 
to  a  point  on  the  north  bank  of  the  Ohio  River.  The  exten- 
sion of  this  line  north  of  the  Ohio  is  known  as  Ellicott's  Line. 
From  this  point  on  the  north  bank  of  the  Ohio  River, 
Hut  chins  started  a  line  westward  as  a  Base  Line.  Accord- 
ing to  Congressional  direction,  he  was  to  lay  off  meridians 
at  intervals  of  six  miles  along  this  east- west  "  Geographer's 
line  "  and  also  parallels  to  this  east-west  line,  every  six 
miles.  Each  of  these  six-mile  squares  was  to  be  divided 
into  thirty-six  square  miles  and  each  of  these  into  **  quar- 
ters." The  large  squares  were  called  "  townships  "  after  the 
New  England  word  ' '  town. ' '  Hutchins  and  his  assistants  laid 
off  only  forty-two  miles  of  the  "  Geographer's  line,"  making 


XIII,  §  196]  LAND  SURVEYING  233 

seven  ranges  of  townships  west  of  the  Pennsylvania  state 
boundary  line.     They  were  frightened  away  by  the  Indians. 

The  Principal  MeridianSj  §  194,  are  run  from  some  initial 
point  selected  with  great  care  and  are  located  in  latitude 
and  longitude  by  astronomical  means.  More  than  thirty- 
two  of  these  principal  meridians  have  been  surveyed  at 
irregular  intervals  and  of  varying  lengths.  Some  of  them 
are  numbered,  while  others  are  named.  The  first  principal 
meridian  (written.  First  P.M.)  is  the  boundary  between 
Indiana  and  Ohio ;  the  second  is  west  of  the  center  of 
Indiana,  extending  the  entire  length  of  the  state ;  the  third 
extends  the  entire  length  of  the  state  of  Illinois  and  through 
the  center ;  the  Tallahassee  principal  meridian,  only  twenty- 
three  miles  long,  runs  directly  through  that  city.  Other 
principal  meridians  are  the  Black  Hills,  Indian,  Louisiana, 
Mount  Diablo,  San  Bernardino,  etc. 

A  Base  Line  is  run  at  right  angles  to  the  principal  meridian 
at  some  point  selected  to  begin  the  survey.  Base  lines  are 
true  geographical  parallels,  and  as  with  the  principal  merid- 
ian, they  are  laid  off  with  great  care. 

The  fourth  P.M.  in  western  Illinois  and  Wisconsin  has 
two  base  lines,  one  at  its  southern  extremity  and  the  other 
on  the  boundary  between  Illinois  and  Wisconsin. 

The  fifth  P.M.  and  its  base  line  carries  the  largest  area 
embraced  in  any  one  system.  This  meridian  extends  from 
the  mouth  of  the  Arkansas  River  northward  through  Ar- 
kansas, Missouri,  and  Iowa.  Its  base  line,  known  some- 
times as  the  Little  Rock  base  line,  passes  a  few  miles  south 
of  that  city  in  Arkansas.  From  this  meridian  and  its  base 
line  all  of  Arkansas,  Missouri,  Iowa,  North  Dakota,  and  the 
major  portions  of  Minnesota  and  South  Dakota  have  been 
surveyed.  This  constitutes  an  area  considerably  larger 
than  that  of  France  and  Great  Britain  and  Ireland  com- 


234  MATHEMATICS  fXIII,  §  196 

bined.  The  most  northern  tier  of  townships  from  this  base 
line  is  near  the  forty-ninth  parallel,  the  boundary  between 
the  United  States  and  Canada,  and  is  numbered  163.  The 
most  northern  township  lies  more  than  a  thousand  miles 
north  of  the  base  from  which  it  was  surveyed.  There  are 
nineteen  tiers  south  of  this  Little  Rock  base  line,  so  that  the 
extreme  length  of  this  area  is  about  1122  miles.  The  most 
eastern  range  from  this  fifth  P.M.  is  number  17  and  its  most 
western  104,  making  an  east  and  west  extent  of  726  miles. 


CHAPTER  XIV 
SIMPLE   MACHINES 

197.  Introduction.  Due  to  wider  and  more  intense  com- 
petition, both  with  man  and  nature,  the  modern  farmer 
cannot  expend  his  strength  in  developing  mere  mechanical 
power  which  may  be  created  by  a  horse  or  a  windmill  or  an 
engine.  He  must  devote  a  larger  and  larger  share  of  his 
time  and  energy  to  mental  activity.  The  result  of  this 
tendency  is  to  increase  the  number  and  the  complexity  of 
his  machines.  Thus  a  knowledge  of  mechanical  principles 
has  become  a  fundamental  necessity  in  order  that  he  may 
properly  handle  and  care  for  machinery. 

198.  Motion.  Force.  Moment  of  a  Force.  Motion  is 
change  of  position.  That  which  produces  motion  or  changes 
it  (or  tends  to  do  so)  is  called  force. 

Force  and  motion  are  transmitted  and  modified  primarily 
by  the  use  of  machines.  Simple  examples  of  machines  are 
a  crowbar  used  to  raise  a  heavy  weight,  a  pair  of  pliers  in 
use,  an  ordinary  lift  pump,  a  pulley,  etc. 

If  a  force  P  (Fig.  158)  is  applied  to  the  end  of  a  lever  work- 
ing over  the  fulcrum  F,  it  tends  to  lift  the  weight  W,  thereby 

producing  a  tendency  to  rotate  about   T „_._4^ ^i ^ 

F  as  an  axis.     This  tendency  to  rotate    tl^^^~"      [ 

is  called   the   moment  of   the  force.  ^^^^^-—-.^^ 

The  perpendicular  distance  from  the  Fia.  158 

axis  .of  rotation  to  the  line  of  action  of  the  force  is  called  the 

moment  arm.    The  moment  of  the  force  is  measured  by  the  force 

times  the  moment  arm  : 

235 


236  MATHEMATICS  PCIV,  §  198 

Moment  of  a  Force  =  ForceXMoment  Arm. 

From  Fig.  158, 

Moment  of  P  about  the  axis  at  F=  PXa, 
Moment  of  W  about  the  axis  at  F=  TFXa' ; 

and  for  equihbrium 

(1)  PXa=WXa\ 

199.  Levers.  In  every  simple  machine  two  forces  are 
involved,  viz.  the  resistance,  or  the  force  to  be  overcome, 
and  the  effort,  or  the  force  necessary  to  overcome  the  resist- 
ance. The  relation  between  these  two  forces  depends  upon 
the  nature  and  the  dimensions  of  the  machine. 

With  levers,  resistance  and  effort  are  inversely  proportional 
to  their  distances  from  the  fulcrum,  i.e. 

W\W'  =  V  \l,     or     Wl=  W'V,     i.e. 

Resistance  (W)  times  its  moment  arm  (I) 

=  effort  (W)   times  its  moment  arm  (V). 


F    I    A 


w 

-l'  >F    A 


B  a\  I       Z\ 

W 


w' 


^ 


w 


w 

Fig.  159.  —  Levers,  showing  Fulcrum  and  Moment  Arm 

The  beam  balance,  the  common  steelyard,  scissors, 
pincers,  the  crowbar,  etc.  are  familiar  examples  of  levers. 

200.  Wheel  and  Axle.  With  the  wheel  and  axle,  resistance 
and  effort  are  inversely  proportional  to  the  radii  of  the  cylinder 
and  wheel  respectively,  i.e. 

W:W'=r'\r,     or     Wr=  W'r'. 

The  windlass  and  the  capstan  are  familiar  examples  of 
the  wheel  and  axle.     Wheels  on  separate  axles,  connected 


XIV,  §  203] 


SIMPLE  MACHINES 


237 


by  belts  or  by  teeth  in  the  circumference 
of  the  wheels,  are  modifications  of  the 
wheel  and  axle. 

201.   A  General  Principle.     Work.     If, 

in  any  simple  machine,  a  body  is  made  to 
move  a  distance  d  directly  against  a  resist- 
ing force  W,  by  an  effort  W  that  moves 
a  distance  d^  in  the  line  of  action  of  W\ 
the  product  W  -  d  is  equal  to  the  product  TF'  •  d\ 
product  is  called  the  work  done. 


w  w 

Fig.  160.  — Wheel 
AND  Axle 

This 


202.  The  Inclined  Plane.  With  the  inclined  plane,  weight 
and  effort  are  directly  proportional  to  the  length  and  the  height 
of  the  incline,  i.e. 

(2)  W'l  W'  =  l:h,  or   Wh=  W'l 

w 


\A 

f 

h 

w 

\ 

FiQ.  161.  —  Inclined  Plane 

203.  The  Screw.  The  screw  is  a  modified  form  of  the 
inclined  plane.  The  effort  is  applied  at  the  end  of  a  lever 
and  the  length  of  the  inclined  plane  is  the  distance  traversed 
by  the  effort  in  one  revolution,  i.e.  2  irl,  where  I  is  the  lever 
arm.  The  height  of  the  incline  is  the  distance  d  between 
two  successive  threads  of  the  screw.  This  distance,  d,  is 
called  the  pitch  of  the  screw. 

Here,  the  weight  and  effort  are  directly  proportional  to  the 
distance  traversed  by  the  effort  in  one  revolution,  and  the  pitch 
of  the  screw,  i.e. 
(3)  W:  W'  =  2TTl:d,     or      Wd  =  2irW'L 


238  MATHEMATICS  PCIV,  §  203 

EXERCISES 

1.  What  effort  is  required  to  lift  a  weight  of  800  lb.  by  a  lever 
whose  weight  arm  is  6  in.  and  whose  effort  arm  is  2  ft.  6  in.  ? 

2.  A  pump  handle  is  3  ft.  8  in.  long  and  works  on  a  pivot  4  in.  from 
the  end  attached  to  the  pump  rod.  What  force  is  applied  to  the  pump 
rod  when  the  end  of  the  handle  is  pushed  down  with  a  force  of  10  lb.  ? 

3.  A  man  wishes  to  lift  a  stone  weighing  1000  lb.  by  means  of  a 
crowbar  6  ft.  long.  If  he  places  the  fulcrum  10  in.  from  the  end  of 
the  bar,  what  force  must  be  exerted  at  the  end  of  the  bar  to  raise  the 
stone?  In  what  direction  must  this  force  be  applied?  If  the  force  is 
applied  at  a  different  angle,  does  it  take  more  or  less  force?  For 
example,  if  the  bar  stands  at  an  angle  of  45°  when  the  force  is  applied, 
in  what  direction  should  the  force  be  applied  ? 

4.  What  force  is  required  to  lift  a  load  of  275  lb.  on  a  wheelbarrow 
if  the  load  is  placed  18  in.  back  of  the  axle  of  the  wheel  and  the  ends  of 
the  handles  are  5  ft.  from  the  axle  ? 

5.  The  crank  to  the  windlass  of  a  well  is  18  in.  long  and  the  cylinder 
upon  which  the  rope  is  wound  is  6  in.  in  diameter.  How  much  force 
is  necessary  to  lift  a  bucket  of  water  weighing  48  lb.,  neglecting  friction? 
How  much  if  the  friction  is  20%  of  the  force  apphed? 

6.  Two  men  working  at  a  capstan  walk  in  a  circle  8  ft.  in  diameter 
and  each  exerts  a  force  of  60  lb.  With  each  complete  turn,  2  ft.  of 
rope  is  pulled  in.     What  is  the  pull  along  the  rope,  neglecting  friction  ? 

7.  A  horse  walking  in  a  circle  15  ft.  in  diameter  moves  a  house  by 
means  of  a  capstan  18  in.  in  diameter.  The  horse  exerts  a  pull  of 
1200  lb.  What  is  the  resistance  of  the  house  if  only  f  of  the  force  exerted 
by  the  horse  is  available  ? 

8.  A  stone  weighing  756  lb.  is  lifted  into  place  in  a  wall  by  means  of 
a  cable  wound  about  an  axle  of  5  in.  radius.  To  this  axle  is  rigidly 
attached  a  toothed  wheel  of  15  in.  radius,  and  this  toothed  wheel  is 
driven  by  another  toothed  wheel  of  4  in.  radius.  This  last  wheel  is 
turned  by  a  crank  18  in.  long.  How  much  force  must  be  applied  to 
the  crank,  neglecting  friction?     How  much,  allowing  20%  friction? 

9.  If  a  barrel  of  cider  weighs  200  lb.,  what  force  is  required  to  roll 
it  up  an  incline  8  ft.  long  into  a  wagon  3  ft.  high?  How  long  would  the 
incline  have  to  be  if  a  man  could  push  only  40  lb.  ? 


XIV,  §  203]  SIMPLE  MACHINES  239 

10.  A  stone  weighs  850  lb.  What  force  is  required  to  pull  it  on 
rollers  up  an  incline  12  ft.  long  and  place  it  upon  a  wagon  30  in.  high? 

11.  What  force  is  required  to  pull  a  loaded  wagon  weighing  3800  lb. 
up  a  slope  30  ft.  high  and  105  ft.  long,  neglecting  friction  ? 

12.  Ice  is  pulled  up  an  incline  50  ft.  long  20  ft.  high  into  an  ice  house. 
What  force,  neglecting  friction,  is  necessary  to  pull  up  a  200-lb.  block? 

13.  A  house  that  is  being  raised  by  jackscrews  exerts  a  pressure 
of  three  tons  on  one  of  them.  If  the  pitch  of  the  screw  is  |  in.  and  it  is 
operated  by  a  lever  2  ft.  long,  what  force  must  be  appUed  at  the  end  of 
the  lever? 

14.  If  the  pitch  of  a  jackscrew  is  f  in.  and  the  lever  arm  30  in., 
what  weight  may  be  hfted  by  it  with  a  force  of  100  lb.  ? 

15.  How  far  will  a  force  of  180  lb.  which  moves  4  ft.  lift  a  weight  of 
960  lb.,  if  one  third  of  the  applied  force  is  consumed  in  overcoming 
friction?  Ans.     6  in. 

16.  What  force,  moving  6  ft.  8  in.,  will  lift  512  lb.  a  distance  of 
1  ft.  3  in.,  if  one  third  of  the  applied  force  is  consumed  in  overcoming 
friction?  Ans.     144  lb. 

17.  CD  is  a  crowbar  6J  ft.  long,  and  is  supported  at  F,  6  in.  from  C. 
How  many  pounds  of  force  must  be  exerted  by  a  man  pressing  down  at 
D  to  raise  a  stone  weighing  1800  lb.  at  A?  To  get  the  maximum  of 
the  force  exerted  at  D,  in  what  direction  should  it  be  applied  ? 

18.  Suppose,  in  Ex.  17,  that  the  crowbar  stands  at  an  angle  of  30° 
with  the  horizontal.  A  man  weighing  150  lb.  sits  at  D  on  the  end  of 
the  bar.     What  is  the  force  communicated  to  the 

bar  at  D?     How  much  will  it  Uft  at  C? 

19.  In  pulling  a  nail  from  a  board,  as  shown  in 
the  accompanying  figure,  find  the  resistance  of  the 
nail  in  starting  if 

P=401b.,  D  =  10",  d  =  ir. 

20.  If  a  gate  weighs  50  pounds  and  is  10'  long, 
what  is  the  pull  on  the  upper  hinge  if  it  is  3'  above 
the  lower  hinge  ? 

[Hint.  The  center  of  mass  is  in  the  center  of  the  gate.  The  system 
of  forces  then  acts  as  a  simple  teeter  board  with  the  fulcrum  on  the 
lower  hinge.]  Ans.    83^  lb. 


240 


MATHEMATICS 


\X1Y,  §  204 


204.  The  Pulley.  The  pulley  consists  of  a  grooved  wheel 
capable  of  revolving  about  an  axis,  fixed  into  a  framework 
called  the  block.  When  the  axis  of  the  pulley 
is  fixed,  the  pulley  is  called  a  fixed  pulley.  It 
has  no  mechanical  advantage,*  its  use  being 
merely  to  change  the  direction  of  the  force. 

If  the  pulley  can  ascend  or  descend,  it  is 
called  a  movable  pulley  and  a  mechanical 
advantage  may  be  gained.  In  Fig.  163,  the 
weight  W  is  supported  by  the  force  P  ap- 
plied to  the  cord  ABDP  fixed  at  A  and 
passing  under  the  pulley.  The  tension  in  the  cord  being 
everywhere  the  same,  we  have 

(4)  P X lever  arm  CB=WX lever  arm  OB, 

But  CB  =  2  OB.     Therefore 


D 

w 
Fig.   163.  — Sim- 
ple  Movable 
Pulley 


(5) 


2' 


or     W=2P; 


and  the  mechanical  advantage  is  2,  since  W/P  =  2. 

205.  System  of  Movable  Pulleys.  In  Fig.  164  is  shown 
a  system  of  three  movable  pulleys  each  hanging  from  a 
fixed  beam  by  a  separate  string. 

The  tension  at  A  in  the  cord  under 
the  lowest  pulley  is  =W/2.  The  ten- 
sion at  B  in  the  cord  under  the  second 
pulley  is  one  half  of  the  first  tension, 
i.e.  at  A.     Thus  the  tension  at  B  is 


W 
2 


W 

22 


Fig.  164.  —  System 
OF  Pulleys 


*  Mechanical  advantage  is  the  ratio  of  the  resisting  force  W  to  the  effort 
force  P,  i.e. 

W  _  distance  P  moves 
P      distance  W  moves 


mechanical  advantage 


XIV,  §  206] 


SIMPLE  MACHINES 


241 


The  tension  in  the  cord  at  C  is  one  half  that  of  the  previous 
one,  etc.  to  the  nth  pulley  for  which  the  tension  in  the  cord 
supporting  it  is  17/2".  Thus,  for  n  pulleys  arranged  as  in 
Fig.  164,  we  have  P  =  TF/2". 

206.   Block  and  Tackle  —  Cord  around  All  Pulleys.     In 

this  system  there  are  two  blocks,  A  and  B,  Fig.  165,  the 
upper  of  which  is  fixed  and  the  lower 
movable,  and  each  containing  a  number 
of  pulleys  or  sheaves,  each  pulley  mov- 
able around  the  axis  of  the  block  in  which 
it  lies.  A  single  cord  is  attached  to  one 
of  the  blocks  and  passes  alternately 
round  the  pulleys  in  the  two  blocks,  the 
portions  of  the  cord  between  successive 
pulleys  being  parallel.  In  the  tackle 
shown  in  Fig.  165,  there  are  three  pul- 
leys in  each  block. 

The  mechanical  advantage  of  the  block  and  tackle  varies 
directly  as  the  number  of  cords  supporting  the  weight. 

For,  as  the  cord  passes  round  all  the  pulleys,  its  tension  is 
the  same  throughout  and  equal  to  P.  Then,  if  n  be  the 
number  of  cords  at  the  lower  block,  nP  will  be  the  resultant 
total  upward  tension  of  the  cords  at  the  lower  block,  which 
must  equal  W.  Therefore 
(6)  nP  =  W,    or 


w  w 

Fig.  165.  — Block 
AND  Tackle 


p=E. 


In  the  system  of  Fig.  165,  there  are  six  cords  supporting 
the  weight  at  the  lower  block.  The  mechanical  advantage 
is  therefore  six.  If  the  lower  block  is  raised  one  foot,  the 
force  at  P  will  move  through  six  feet,  and  a  given  force  at  P, 
say  100  lb.,  will  therefore  raise  a  weight  at  W  six  times  as 
great,  or  600  lb. 


242  MATHEMATICS  [XIV,  §  206 

EXERCISES 

1.  What  force  is  necessary  to  raise  a  weight  of  450  lb.  by  an  arrange- 
ment of  six  pulleys  in  a  block  and  tackle  ? 

Arrange  a  block  and  tackle  to  lift  500  lb.  if  only  50  lb.  is  available  as 
the  power. 

2.  Find  the  force  which  will  support  a  weight  of  720  lb.  with  three 
movable  pulleys,  arranged  as  in  Fig.  164. 

If  50  lb.  is  the  available  force,  how  many  movable  pulleys  arranged 
as  in  Fig.  164  are  necessary  to  lift  750  lb.  ?     1550  lb.  ? 

3.  A  man  weighing  145  lb.  raises  a  weight  of  448  lb.  by  a  system  of 
four  movable  pulleys  arranged  as  in  Fig.  164.  What  is  his  pressure  on 
the  ground  as  he  raises  the  weight  ?  Ans.     11 7  lb . 

4.  How  many  movable  pulleys  are  required  in  a  block  and  tackle 
if  42  lb.  is  available  with  which  to  raise  336  lb.  ? 

207.  Strength  of  Materials.  In  all  farm  buildings  and 
farm  implements  the  materials  used  are  subjected  to  certain 
stresses  and  strains.  The  materials  may  be  classified  ac- 
cording to  their  physical  structure  under  three  headings : 
(1)  wood ;  (2)  metal ;  (3)  stone,  brick,  cement,  mortar, 
concrete. 

The  kind  of  material  used,  and  its  design,  must  meet  the 
demands  made  upon  it  without  breaking.  Two  prime  con- 
siderations in  the  use  of  materials  for  machines  and  buildings 
are  the  dze  and  shape  of  the  material,  as  when  used  as  a  joist 
or  a  reaper  tongue ;  and  the  manner  in  which  it  is  loaded. 

The  strength  of  a  beam  is  measured  by  the  load  it  can 
safely  withstand.  The  safe  load  which  a  beam  will  carry 
varies  as  each  of  the  following  factors. 

(1)  Directly  as  the  working  fiber  strength  of  its  material. 
Each  kind  of  material  thus  is  given  a  rating  or  a  constant. 
This  constant  provides  also  for  a  reasonable  factor  of  safety. 
In  the  table  presented  below,  §  208,  this  factor  of  safety  is 
six.  This  means  merely  that  the  safe  load  is  one  sixth  of 
the  breaking  load. 


KIV,  §  208] 


SIMPLE  MACHINES 


243 


(2)  Directly  as  the  breadth  times  the  square  of  the  depth  for 
beams  with  rectangular  cross-section.  Wooden  beams  are 
usually  of  this  form. 

(3)  Inversely  as  the  length. 
Thus  we  may  write 

(7)  safe  load  =  constant  (breadth)  (depth)^  ^ 

length 

208.  Safe  Loads  for  Beams  under  Bending  Stress.  There 
are  at  least  four  common  ways  of  loading  beams.  We 
present  the  formulas  for  safe  loads  in  these  four  cases,  with 
the  following  notation. 

*S  =  safe  load  in  pounds,  d= depth  in  inches, 

6  =  breadth  in  inches,  L  =  span  in  feet, 

A  =  constant  depending  on  the  material. 

The  value  of  A  is  given  in  the  following  table  for  some  com- 
mon materials. 


Material 

Const.  A 

Material 

Const.  A 

Yellow  Pine    .     .     . 
White  Pine     .     .     . 
Hemlock     .... 
Chestnut    .... 

100 
60 
50 
55 

White  Oak       .     .     . 
Castiron      .... 

Steel       

Limestone   .... 

75 
167 

889 
8 

Case  I.    Beam  supported  at  one  end 
and  loaded  at  the  other  (Fig.  166). 


(8) 


bd'A 
4  L 


Case  II.    Beam  supported  at  one  end  M 


and  uniformly  loaded  (Fig.  167). 
bd^A 


(9) 


S  = 


2L 


FiQ.  167 


244 


MATHEMATICS 


CSIV,  §  208 


Case  III.  Beam  supported 
at  both  ends  and  loaded  in  the 
middle  (Fig.  168). 

L 


(10) 


Case  IV.  Beam  supported 
at  both  ends  and  uniformly 
loaded  (Fig.  169). 

(11)  s=2Mi. 


Fig.  168 


oooooomonnon 


Fig.  109 


EXERCISES 

1.  A  white  oak  timber  2  in.  in  width  is  to  be  used  to  support  a  hay- 
fork and  carrier  at  the  end  of  a  barn.  How  deep  must  the  timber  be 
to  carry  safely  a  load  of  1200  lb.  applied  3  feet  from  the  single  support? 

Ans.  (i  =  9.8"  =  10"  piece  to  be  used. 

2.  Compare  a  2"  XlO"  and  a  3"  X8"  timber  as  to  cost  and  carrying 
capacity  when  used  as  a  beam,  edgewise.  Do  the  same  when  used 
flatwise. 

3.  A  steel  beam  1|"  wide  is  to  be  used  to  support  a  hayfork  and 
carrier  at  the  end  of  a  barn.  How  deep  must  it  be  to  carry  safely  a 
load  of  1200  lb.  applied  3  ft.  from  the  single  support? 

4.  What  should  be  the  maximum  distance  between  supports  for  a 
2"  XlO"  hemlock  beam  to  safely  sustain  a  middle  load  of  1000  lb.  ? 

5.  What  length  of  beam  could  be  used  in  Ex.  4,  if  the  load  of  1000  lb. 
is  uniformly  distributed? 

6.  Find  the  ratio  of  the  safe  loads  of  two  beams  of  wood,  one  being 
10'  long,  3"Xl2"  in  cross-section  and  having  its  load  in  the  middle, 
and  the  other  8'  long,  2"  X8"  in  cross-section  with  its  load  uniformly 
distributed. 

7.  What  should  be  the  breadth  and  depth  of  a  yellow  pine  rectangular 
beam  20'  between  the  two  end  supports  to  carry  safely  a  load  of  1500  lb. 
at  its  center? 

[Hint.  The  selection  must  be  governed  by  commercial  sizes,  as 
the  dimensions  of  wooden  beams  are  not  fractional.] 


XIV,  §  209]  SIMPLE  MACHINES  245 

209.  Work.  Work,  in  its  scientific  sense,  is  done  only  by 
a  force  in  overcoming  resistance.  The  mere  push  or  pull  of 
the  force  is  not  work.  In  order  that  work  may  be  done, 
the  point  of  application  of  the  force  must  move. 

The  work  done  is  equal  to  the  force  applied  multiplied  hy 
the  distance  through  which  it  acts.     (See  §  201.)      That  is, 

(12)  Work=  ForceX  Distance. 

To  compute  the  work  done  multiply  the  force  in  pounds  by 
the  distance  in  feet.  The  product  is  expressed  in  foot-pounds. 
A  foot-pound  of  work  is  said  to  be  done  when  a  force  of  one 
pound  moves  its  point  of  application  through  a  distance  of 
one  foot  in  the  direction  of  the  force. 

Example.  How  much  work  is  done  when  a  force  of  20  pounds  moves 
its  point  of  application  40  feet  in  the  direction  of  the  force  ? 

A  force  of  20  pounds  acting  through  40  feet  is  the  equivalent  of 
800  pounds  acting  through  1  foot,  i.e.  800  foot-pounds  of  work  is  said 
to  be  done. 

EXERCISES 

1.  How  much  work  is  done  by  a  man  weighing  150  pounds  in  climb- 
ing to  the  top  of  a  barn  40  feet  high?  How  much  is  done  by  a  man 
weighing  120  pounds  in  walking  up  a  hill  40  feet  high?  How  much 
by  a  horse  weighing  1200  pounds  in  walking  up  a  hill  100  feet  high? 

2.  Twenty  pounds  will  pull  a  certain  cart  on  a  level  road.  How 
much  work  is  done  by  a  boy  in  pulling  it  a  mile  on  this  road  ? 

3.  The  draft  of  a  certain  plow  {i.e.  the  force  required  to  pull  it)  is  500 
lb.     How  much  work  is  done  in  plowing  a  furrow  100  rods  long? 

How  much  work  is  done  by  this  same  plow  in  plowing  a  piece  of 
land  60  rods  long  and  6  rods  wide  if  the  furrow  slice  is  12  inches  wide? 

4.  What  is  the  draft  of  a  wagon,  if  400,000  foot-pounds  of  work  is 
done  by  a  horse  in  drawing  it  a  mile? 

5.  A  baseball  weighing  5x\  oz.  was  dropped  from  one  of  the  windows 
at  the  top  of  Washington's  monument  and  caught  4  ft.  from  the  ground 
by  a  professional  catcher.  Compute  the  work  done  by  the  ball,  assum- 
ing the  window  to  be  545  feet  from  the  ground. 


246  MATHEMATICS  fXIV,  §  209 

6.  A  certain  gas  engine  has  a  4|-inch  bore,  a  5|-inch  stroke,  and 
makes  378  strokes  per  minute.  How  much  work  is  done  per  minute 
if  the  average  pressure  against  the  piston  head  is  15  lb.  per  square  inch? 

210.  Horse-power.  If  a  man  cuts  100  shocks  of  corn  in 
a  day  of  10  hours,  he  does  it  at  the  rate  of  10  shocks  per 
hour.  A  horse  which  covers  21  miles  in  3  hours  has  made 
the  distance  at  the  rate  of  7  miles  per  hour.  A  belt  which 
moves  50  feet  per  second  has  a  rate  of  3000  feet  per  minute. 
A  horse  which  trots  a  mile  in  2  minutes  is  covering  the 
distance  at  the  rate  of  44  feet  per  second.  If  a  force  of  20 
pounds  acts  through  a  distance  of  300  feet  for  5  minutes, 
it  does  6000  foot-pounds  of  work  but  does  it  at  the  rate  of 
1200  foot-pounds  per  minute.  These  illustrations  will  make 
clear  the  following  definition. 

A  horse-power  is  the  unit  for  the  rate  at  which  work  is 
done.  This  unit  is  33,000  foot-pounds  per  minute,  i.e.  one 
horse-power  means  that  a  body  is  moved  by  a  force  of  33,000 
pounds  in  the  direction  of  the  force  one  foot  in  one  minute. 

The  value  of  this  unit  was  adopted  by  James  Watt  as 
the  result  of  experiments  with  strong  dray  horses  in  England. 
He  was  aware,  however,  that  it  is  in  excess  of  what  can  be 
done  by  an  average  horse  over  a  full  working  day. 

211.  Working  Day  for  a  Horse.  Tests  show  that  a 
steady  pull  equal  to  one  tenth  of  the  weight  of  the  horse, 
for  a  ten-hour  daily  service,  at  the  walking  rate  of  2.5  miles 
per  hour,  is  an  average  of  effective  service. 

One  horse-power  is  developed  by  a  horse  pulling  with  a 
force  of  150  pounds  2|  miles  per  hour,  since  this  means 

5280  X2i 


60 


220  feet  per  minute. 


and  150  pounds  220  feet  per  minute  is  33,000  foot-pounds  per 
minute,  or  one  horse-power. 


XIV,  §  211]  SIMPLE   MACHINES  247 

However,  this  power  will  not  be  developed  by  the  average 
horse  at  the  present  time,  since  to  draw  150  pounds  2 J  miles 
per  hour  and  10  hours  per  day,  a  horse  should  weigh  from 
1400  to  1800  pounds.  This  is  larger  than  the  ordinary  horse 
used  in  America. 

The  average  horse  will  pull  about  120  pounds  during  the 
average  working  day. 

Hence,  the  average  horse  will  develop 

5280X2^X120  ^  25,400  foot-pounds  per  minute 
60 

=  ^  horse-power. 

EXERCISES 

1.  A  horse  draws  a  load  having  a  draft  of  150  pounds  at  the  rate 
of  3  miles  per  hour.  What  horse-power  is  he  developing?  At  this 
rate  should  this  horse  be  required  to  work  full  time  ? 

2.  How  much  power  is  a  horse  developing  when  walking  4  miles 
per  hour  and  pulling  75  pounds  on  his  traces?     120  pounds? 

3.  A  horse  walks  2.5  miles  per  hour  and  pulls  120  pounds  on  his 
traces.  How  fast  should  he  walk  to  develop  the  same  power  when 
the  draft  is  increased  to  160  pounds? 

4.  How  much  power  does  a  1600-pound  horse  develop  when  walk- 
ing 2.5  miles  per  hour  if  he  pulls  one  tenth  of  his  weight  ?  A  1200-pound 
horse?     A  1000-pound  horse? 

5.  If  a  horse  weighs  1300  pounds  and  is  pulling  one  tenth  of  its 
weight,  how  fast  should  it  walk  to  develop  one  horse-power?  If  this 
horse  walks  3  miles  per  hour,  what  horse-power  is  developed  ? 

6.  What  should  be  the  weight  of  a  horse  which  develops  1.3  horse- 
power in  walking  3  miles  per  hour  and  pulhng  about  one  tenth  of  its 
weight? 

7.  What  power  should  be  provided,  allowing  25%  loss  of  power 
due  to  friction,  to  unload  hay  at  the  rate  of  one  half  ton  per  minute  if 
raised  to  a  height  of  30  feet? 

8.  How  much  power  is  necessary,  if  f  of  it  is  available,  to  raise 
corn  in  an  elevator  to  a  height  of  50  feet  at  the  rate  of  3000  bushels 
per  hour?   (1  bushel  of  corn  =  70  pounds.) 


248  MATHEMATICS  [XIV,  §  211 

9.  In  filling  a  water  tank  60  feet  high,  the  water  is  raised  at  the 
rate  of  200  gallons  per  minute.  What  horse-power  engine  should  be 
provided,  granting  that  only  |  of  the  power  developed  is  available, 
and  allowing  in  addition  a  10%  margin  of  safety  for  extraordinary  pulls? 

10.  The  maximum  load  on  a  hayfork  is  800  pounds.  What  horse- 
power is  required  to  raise  this  load  40  feet  in  1  minute?  How  many 
horses  each  weighing  1000  pounds  would  be  required  to  furnish  this 
power  if  they  walk  3  miles  per  hour  and  if  only  |  of  the  power  applied 
is  available? 

11.  An  elevator  has  a  capacity  of  1500  pounds.  What  power  is 
necessary  to  operate  this  elevator  through  200  feet  in  2  minutes  ? 

12.  Tests  have  shown  that  for  the  old-fashioned  type  of  walking 
plow  a  furrow  slice  8  Xl3  inches  requires  a  draft  of  673.3  pounds.     Find 

(1)  The  horse-power  required  to  draw  this  plow  2  miles  per  hour. 

(2)  The  weight  of  each  of  three  horses  to  pull  this  plow  2  miles  per 
hour  and  10  hours  per  day. 

13.  For  a  modern  walking  plow,  a  furrow  slice  8  Xl3  inches  requires 
about  400  pounds  draft.     Find 

(1)  The  rate  at  which  the  three  horses  of  the  previous  example  may 
walk  in  drawing  this  plow  10  hours  per  day. 

(2)  The  horse-power  required  to  draw  this  modern  plow  2|  miles 
an  hour. 

212.  Draft.  Draft  is  the  amount  of  force  required  to 
move  an  object.  The  fundamental  principle  involved  in 
overcoming  resistance  or  in  doing  work  is  as  follows. 

The  force  doing  the  work  when  multiplied  by  the  distance 
through  which  it  moves  is  equal  to  the  resistance  multiplied  by 
the  distance  through  which  it  moves.  (See  §  201.)  That  is,  in 
the  case  of  raising  a  weight  against  the  force  of  gravity, 

ForceXForce  Distance  =  WeightX  Weight  Distance. 

The  amount  of  increase  or  decrease  of  draft  up  or  down  a 
grade  does  not  depend  upon  the  draft  on  the  level  but  de- 
pends only  upon  the  weight  of  the  load  and  upon  the  slope 
of  the  grade.     For  grades  less  than  15%  the  horizontal 


XIV,  §  212]  SIMPLE   MACHINES  249 

distance  is  approximately  equal  to  the  actual  length  along 
the  grade,  i.e.  the  hypotenuse  is  approximately  equal  to 
the  base  of  the  right  triangle. 

Example.  What  force  is  required  to  pull  a  load  of  60  pounds 
along  a  10%  grade,  neglecting  ^ 

friction?  R^ 

A  10%  grade  means  10  feet 
rise  in  100  feet.  The  force  P 
moves  horizontally  100  feet 
and  the  weight  rises  10  feet 
since 

/2Q  =  60    pounds,    BC  =  \0  ^''''  ^'^^ 

feet,  and  AB  =  100  feet  =  AC  approximately  for  grades  of  low  per  cent. 

Therefore 

PX100  =  60X10,     or     P  =  6  pounds. 

Also,  from  similar  triangles 

QP^BC      .        gP^ll 
QR     AB'      '  '      60      100' 

and  therefore  QP,  parallel  to  the  incline,  is  equal  to  6  pounds.  This 
is  the  force  necessary  to  keep  the  object  from  sliding  down  the  inchne, 
neglecting  friction. 

But  this  amounts  finally  to  merely  multiplying  the  load  by  the 
grade  per  cent  to  obtain  the  increased  or  decreased  draft  up  or  down  a 
grade.     Hence  we  have  the  following  rule. 

For  draft  up  a  grade,  add  to  the  draft  on  the  level  a  quantity 
obtained  by  multiplying  the  load  by  the  grade  per  cent. 

For  draft  down  a  grade,  subtract  from  the  draft  on  the  level 
the  grade  per  cent  of  the  total  load. 

Example.     The  draft  of  4000  pounds  on  a  certain  level  road  is 
250  poimds.     What  is  the  draft  on  a  3%  grade? 
3%  of  4000  lb.  =  120  lb. 

Therefore  250  lb. +  120  lb.  =370  lb.,  the  draft  up  a  3%  grade, 
and  250  lb.  - 120  lb.  =  130  lb.,  the  draft  dovm  a  3%  grade. 


250  MATHEMATICS  \X1Y,  §  212 

EXERCISES 

1.  A  total  load  of  3700  pounds  has  a  draft  of  108  pounds  per  ton  on 
a  certain  level  road.  What  is  the  draft  up  a  3%  grade?  Down  a 
5%  grade?     Up  a  13%  grade?     Down  a  15%  grade? 

2.  A  total  load  of  4500  pounds  has  a  draft  of  560  pounds  up  a  3% 
grade.  What  is  the  draft  per  ton  on  the  level?  What  is  the  draft 
down  a  5%  grade  for  this  same  load?     Up  a  12%  grade? 

3.  A  load  of  2600  pounds  has  a  draft  of  155  lb.  on  a  level  road. 
What  is  the  per  cent  grade  down  which  the  draft  is  zero  ?  What  would 
be  the  draft  up  this  grade?  How  many  degrees  in  the  angle  of  eleva- 
tion for  this  grade  ? 

4.  At  what  rate  may  a  team  of  1350-pound  horses  draw  a  total  load 
of  two  tons  on  a  level  road,  if  the  total  draft  is  165  lb.  ? 

5.  What  should  be  the  size  of  a  two-horse  team  to  draw  a  load  of 
2  tons  up  a  4%  grade  3  mi.  per  hour,  if  the  draft  on  the  level  is  225  lb.  ? 


CHAPTER  XV 

COMPOSITION   AND   RESOLUTION   OF   FORCES 

213.  Graphic   Representation   of   Forces.     See    §  5.    A 

force  to  be  completely  described  must  have  a  point  of  ap- 
plication, a  direction,  and  a  magnitude. 

Since  a  segment  of  a  line  may  have  all  of  these,  it  may  be 
used  to  represent  a  force. 

In  Fig.  171,  let  P  be  a  force  applied  at  0  and  in  the 
direction  OP.  Lay  off  on  OP  sl  number  of  units  equal 
to  the  magnitude  of  the  force.     When  y 

the    arrowhead   is   affixed    to   indicate 
the  sense  in   the   line   of   motion,  the 
force   is   completely   determined.     For  _ 
example,  if  a  force  of  10  lb.  is  repre- 
sented on  a  drawing  by  1",  then  30  ^^^'-  ^'^^ 
lb.  on  the  same  drawing  should  be  represented  by  3". 

214.  Composition  of  Motions  and  of  Forces.  Parallelo- 
gram of  Forces. 

The  resultant  of  two  or  more  forces  is  that  force  which 
singly  will  produce  the  same  effect  as  the  forces  themselves 
when  acting  together. 

The  components  of  the  resultant  are  the  individual  forces 
which  would  together  produce  the  same  effect  as  the  resultant. 

The  composition  of  motions,  or  of  forces,  or  of  velocities, 
is  merely  the  process  of  finding  their  resultant.  The  follow- 
ing cases  arise. 

251 


252 


MATHEMATICS 


[XV,  §  214 


First.  //  two  or  more  forces  act  in  the  same  straight  line, 
the  resultant  force  is  equal  to  their  algebraic  sum,  and  acts  in 
the  same  line.  If  the  forces  are  not  all  in  the  same  sense, 
either  sense  may  be  selected  as  positive  and  the  other  as 
negative. 

Second.  //  the  forces  do  not  act  in  the  same  straight  line, 
the  parallelogram  of  forces  is  used  to  find  their  resultant,  as 
follows. 

When  a  body  is  acted  upon  by  two  forces,  we  draw  two 
line-segments,  AD  and  AB,  Fig.  172,  to  represent  the 
forces,  as  in  §  213.  If  we  com- 
plete the  parallelogram  DABC 
and  draw  the  diagonal  AC,  that 
diagonal  represents  the  resultant 
of  the  two  given  forces  both  in 
direction  and  in  magnitude. 

Velocities  may  be  combined  to 
find  a  resultant  velocity  in  precisely  the  same  manner. 

215.  Computation  of  the  Resultant  of  Two  Given  Forces. 
Let  the  two  forces  be  Fi,  F2  and  be  represented  hy  AB  and 
AD,  respectively  (Fig. 
173).  Complete  the  par- 
allelogram DABC,  as  in 
§214,  and  let  i2  =  AC  be 
the  resultant.  Let  0  de- 
note the  angle  between 
the  forces,  and  let  a  de- 
note the  angle  between  Fi  and  the  resultant, 
on  AB.    Then  we  have 

(1)  BE =F2  cos  e, 

(2)  ^C  =  i^2sin^,  I 

(3)  AE=Fi-\-F2C0se, 


A  B 

Fig.  172.  —  Parallelogram 
OF  Forces 


Project  BC 


XV,  §  215]  COMPOSITION  OF  FORCES  253 

and 


(4)  R=  ylAE'+EC\ 

(5)  tana=^,    and    R=  ^^ 


AE  sin  a 

From  these  formulas,  the  resultant  R  and  its  direction  a 
may  be  computed. 

EXERCISES 

1.  Three  forces  of  5  lb.,  3  lb.,  and  2  lb.,  respectively,  act  upon  a 
point  in  the  same  straight  line  in  one  sense,  and  two  other  forces  of 
8  lb.  and  9  lb.  act  in  the  same  line  in  the  opposite  sense.  With  what 
force  and  in  which  direction  will  the  point  move? 

Draw  the  conditions  to  the  scale  1  lb.  =^". 

2.  Two  forces  whose  magnitudes  are  as  3  to  4  acting  on  a  point 
at  right  angles  to  each  other  produce  a  resultant  of  30  lb.  What  are 
the  component  forces  ?     What  is  the  direction  of  the  resultant  ? 

Ans.     18  lb.  and  24  lb. 

3.  Let  ABC  be  a  triangle,  and  D  the  middle  point  of  the  side  BC. 
If  the  three  forces,  represented  in  magnitude  and  direction  by  AB, 
AC,  and  AD,  act  upon  the  point  A,  find  the  direction  and  magnitude  of 
the  resultant.  Ans.     3  AD  in  the  direction  AD. 

4.  What  single  force  would  be  sufficient  to  counteract  a  vertical 
force  of  5  pounds  and  a  horizontal  force  of  12  pounds  acting  on  a  body 
at  the  same  instant?     What  would  be  the  direction  of  this  resultant? 

Ans.     13  pounds,  22°  37'  with  the  horizontal. 

5.  Find  the  magnitude  and  the  direction  of  the  resultant  of  two 

forces  of  30  pounds  and  40  pounds  acting  on  a  body  at  an  angle  of  60° 

with  each  other.  •    ^ 

■Dj -^c 

[Hint.     In    the   accompanying   figure,    find         '  /  ^y"^/^'* 

BE  and  EC.     Then  CA  may  be  found  from  the        y  o    ^^^^  ^/  ! 

right  triangle  AEC]  ly^^Kr  M      ' 

^  OQ  -^j  jn 

6.  A  river  is  flowing  at  the  rate  of  3  mi.  per 
hour.     How  should  a  man,  rowing  a  boat  5  miles 

per  hour,  point  the  boat  up  the  river  in  order  that  his  course  may  be 
directly  across  the  river?  What  would  be  his  rate  directly  across? 
What  is  the  tiirm  across  if  the  river  is  2  mi.  wide? 


254  MATHEMATICS  [XV,  §  215 

7.  A  man  rows  a  boat  across  a  river  at  the  rate  of  3.5  miles  per  hour. 
The  river  flows  at  the  rate  of  4.8  miles  per  hour.  Find  the  speed  of  the 
boat.  Where  will  it  land  along  the  opposite  bank  if  the  river  is  2  miles 
wide? 

8.  Given  Fi  =  37.6,  ^2  =  58.7,  ^  =  58°  20'  (Fig.  173,  §215),  find  R 
and  a. 

9.  Given  i^i  =  37.6,  ^2  =  58.7,  ^  =  71°  40',  find  R  and  a. 

10.  Given  Fi  =  62.7,  ^2  =  46.8,  ^  =  90°,  find  R  and  a. 

11.  From  an  auto  going  25  mi.  per  hour  a  man  throws  a  stone  at 
the  rate  of  60  ft.  per  second  and  at  an  angle  of  45°  with  the  line  of 
motion  of  the  auto.     Find  the  speed  and  direction  of  the  stone. 

12.  Two  forces  of  4  lb.  and  3^2  lb.  act  at  an  angle  of  45°,  and  a 
third  force  of  V42  lb.  acts  at  right  angles  to  their  plane  at  the  same 
point.     Find  their  resultant.  Ans.     10  lb. 

216.  Resolution  of  Forces.  By  the  resolution  of  forces 
is  meant  the  process  of  finding  the  components  of  given  forces 
in  two  given  directions.  We  saw  in§  215  that  two  forces,  Fi 
and  F2,  acting  at  a  point  are  equivalent  to  a  single  force,  R, 
the  resultant.  It  is  evident  then  that  the  single  force,  R, 
acting  along  the  diagonal  AC  of  the  parallelogram.  Fig.  173, 
can  be  replaced  by  the  two  forces,  Fi  and  F2,  represented  in 
magnitude  and  direction  by  two  adjacent  sides  of  the  par- 
allelogram. 

The  given  total  force  AC  may  be  used  as  the  diagonal  for 
an  infinite  number  of  parallelograms.  For  we  can  draw 
through  A  two  lines  AD  and  AB  in  any  given  directions. 
It  follows  that  a  single  force  may  be  resolved  into  com- 
ponents in  any  two  given  directions. 

217.  Rectangular  Components  of  a  Force.  The  most 
convenient  components  into  which  a  force  can  be  resolved 
are  those  whose  directions  are  at  right  angles  to  each  other. 
Thus  let  OX  and  OF  be  any  two  lines  at  right  angles  to 


XV,  §  217]         COMPOSITION  OF  FORCES  255 

each  other  and  P,  Fig.  175,  any  force  acting  at  0  in  the 
plane  XOY.  Completing  the  rectangle  OMPN,  we  find 
the  components  along  the  two  axes : 

(6)  a:-component  =  OM  =  P  cos  a, 

(7)  i/-component  =  ON  =  P  sin  a, 

where  a  is  the  angle  which  the  di- 
rection of  P  makes  with  the  x-axis.    _ 
The  X-  and  the  ^/-components  of 

a  force  are  called  its   rectangular     ^^°-  175.  —  Rectangular 

Components  of  a  Force 
components.     They  are  usually  the 

horizontal  and  the  vertical  components.     See  also  §  5. 

Unless  otherwise  expressed,  the  components  of  a  force 
will  mean  its  rectangular  components  on  a  horizontal  line 
and  a  vertical  line. 

Velocities  may  be  resolved  into  component  velocities  in 
precisely  the  same  manner. 

EXERCISES 

1.  The  tension  in  each  of  four  traces  by  which  two  horses  are  drawing 
a  plow  is  50  lb.  What  are  the  horizontal  and  vertical  components  of 
the  total  force  moving  the  plow  if  the  traces  make  an  angle  of  14°  with 
the  horizontal  ? 

2.  A  ball  rolls  along  the  diagonal  of  a  6'  Xl8'  wagon  bed  at  the  rate 
of  10  feet  per  second.  What  is  the  speed  of  the  ball  when  the  wagon 
is  moving  forward  at  the  rate  of  12  feet  per  second? 

3.  Sleet  is  falling  with  a  vertical  speed  of  16  feet  per  second.  What 
is  the  velocity  of  the  wind  when  the  sleet  falls  at  an  angle  of  60°  with  the 
horizontal  ?  What  is  the  wind's  velocity  when  the  sleet  falls  at  an  angle 
of  45°  with  the  horizontal  ? 

Ans.     6.3  mi.  per  hour;  11  mi.  per  hour,  approximately. 

4.  What  must  be  the  tension  in  a  rope  to  support  on  a  smooth  ring 
a  pig  weighing  100  lb.  between  two  posts  12  feet  apart  if  the  rope  is 
allowed  to  sag  3  feet  at  the  center?  What  is  the  tension  when  the  sag 
is  2  feet?  Ans.     112  lb.,  158  lb. 


256 


MATHEMATICS 


[XV,  §  217 


5.  A  weight  W  lies  on  a  smooth  inclined  plane  which  makes  an 
angle  cc  with  the  horizontal.  Show  that  the  components  of  W  along 
and  perpendicular  to  the  plane  are 

P  =  TF  sin  a,     R  =  W  cos  a, 


where  R  is  the  reaction  against  the  plane  and  P  is  the  force  acting 
parallel  to  the  plane  to  keep  the  weight  from  slipping. 

6.  A  weight  of  10  lb.  is  held  in  position  on  a  smooth  inclined  plane 
(inclination  a)  by  a  force  of  2  lb.  acting  up  the  plane  and  by  a  force 
acting  horizontally  on  the  weight.  Determine  this  horizontal  force  if 
sina  =  f. 

[Hint.  Consider  the  components  of  all  of  the  forces  in  the  direction 
of  the  plane.]  Ans.     5  lb. 

7.  A  farmer  sets  three  posts  in  the  ground  so  as  to  form  an  equi- 
lateral triangle.  A  rope  is  passed  around  them  the  tension  in  which 
is  6  lb.     Find  the  pressure  on  the  posts.  Ans.     6 V3  lb. 

8.  A  barn  is  equipped  with  a  hayfork,  pulleys,  etc.  The  rope  lead- 
ing from  the  roof  pulley  to  the  floor  pulley  makes  an  angle  of  70°  with 
the  horizontal.  In  drawing  up  hay  the  tension  in  the 
rope  is  800  lb.  and  the  rope  beyond  the  floor  pulley 
makes  an  angle  of  10°  with  the  horizontal.  Find  the 
vertical  force  necessary  to  hold  the  floor  pulley  to  its 
fastening.  Use  scale  diagrams  as  an  approximate 
check.  Consider  the  entire  rope  at  all  times  in  the 
same  vertical  plane.     Study  the  following  cases : 

(a)  When  the  draft  end  of  the  rope  is  to  the  left  of  the  floor  pulley 
and  10°  below  the  horizontal. 

(b)  When  it  is  to  the  left  of  the  floor  pulley  but  10°  above  the  hori- 
zontal. 


Fig.   177 


XV,  §  217]         COMPOSITION  OF  FORCES 


257 


(c)  When  it  is  to  the  right  of  the  floor  pulley  and  10**  above  the 
horizontal. 

9.  A  wind  is  blowing  nearly  from  the  west,  as  shown  in  the  accom- 
panying figure,  against  the  sail,  AB,  of  a  vessel  going  due  north.  What 
part  of  this  wind  force,  WC,  tends  to  drive  the  vessel  north  ? 

We  may  regard  the  wind  force,  WC,  as  the  resultant  of  two  forces, 
DC,  perpendicular  to  the  sail  and  tending  to  drive  the  vessel  nearly 
northeast  and  WD  parallel  to  it.  But  DC  may 
be  resolved  into  DE  and  EC,  one  of  which  tends 
to  push  the  vessel  sidewise,  the  other  propels  it 
northward.  The  side  thrust  is  largely  counter- 
acted by  the  water. 

10.  Show  how  the  rigging  of  the  vessel  in  the 
previous  exercise  may  be  shifted  so  that  the  same 
wind  will  propel  the  vessel  south. 

Will  the  component  propelling  the  vessel  south  be  equal  to  the  one 
propeUing  it  north? 

11.  Two  equal  rafters  support  a  weight  W  at  their  upper  end.  Find 
the  compression  in  each.  Let  the  length  of  each  ^ 
rafter  be  a  and  the  horizontal  distance  between 
their  lower  ends  be  2  h,  Fig.  179. 

Let  BW  represent  in  direction  and  magnitude 
the  weight  W.  From  W  draw  WH  parallel  to  AB 
and  WK  parallel  to  EB.  Then  BK  and  BH  rep- 
resent the  components  of  the  force  BW  along  the 
rafters  BA  and  BE  (§  216).  Draw  CE  parallel  to 
BL  and  extend  AB  to  C.     From  the  similar  triangles  BWK  and  CEB, 

BK  a  .' .      r.T^  aW 

W 


2Va2-62 


i.e.     BK  = 


2Va2-62 


Find  also  the  tension  in  the  tie-rod  along  b,  by  taking  the  component 
of  BK  on  the  line  AE. 

12.  Find  the  thrust  in  the  equal  rafters  and  the  tension  in  the  tie- 
rod  for  rafters  10  ft.  long  in  a  barn  16  ft.  wide  due  to  a  total  weight  of 
1001b. 

What  are  the  thrust  and  the  tension  for  rafters  9  ft.  long  in  a  barn 
16  ft.  wide? 

What  is  the  efifect  of  short  rafters? 


258 


MATHEMATICS 


[XV,  §  218 


218.  Triangle  of  Forces.  The  resultant  of  two  forces 
F]  and  F2,  Fig.  180,  may  be  obtained  more  easily  by  drawing 
AB  to  represent  the  magnitude  and  direc- 
tion of  one  force  and  BC,  similarly,  for  the 
other  force.  Then  AC  represents  the  re- 
sultant. The  force  CA  (in  direction  from 
Cto  A)  is  opposite  in  direction  but  equal 
in  magnitude  to  the  resultant  of  Fi  and  F2. 

//  three  forces  he  represented  in  magnitude  and  direction 
by  the  sides  of  a  triangle,  taken  in  order,  they  will  he  in 
equilibrium. 


219.  The  Simple  Jib  Crane.  For  moving  heavy  weights, 
a  machine  called  a  crane  is  much  used.  It  consists  of  a  crane 
post  AB,  Si  tie-rod  BC,  a  jib  CA.  When  loaded  with  a 
weight  W  attached  at  C,  the  point  C  is  in  equilibrium  under 
the  action  of  the  three  forces 

(a)  the  weight  W, 

(6)  the  pull  in  the  tie-rod, 

(c)   the  thrust  in  the  jib. 

These  form  a  triangle  of  forces  in  which  any  one  of  the  forces 
is  the  resultant  of  the  other  two,  and  they  may  be  repre- 
sented on  a  convenient  scale  by  the  three  sides  of  a  triangle, 
taken  in  order.  In  Fig.  181, 
CW  represents  the  weight  to 
scale.  Draw  WD  parallel  to 
AC  and  extend  BC  to  D. 
Then  on  the  same  scale  CD  n—^ 
represents  the  pull  in  the  tie-  ^— ^ 
rod  and  WD  the  thrust  in  the 
jib.  The  sense  of  the  forces  is  obtained  by  following 
around  the  triangle  in  order,  from  C  to  IF  to  i)  to  C. 


Fig.  181.  —  The  Jib  Crank 


XV,  §  219]  COMPOSITION  OF  FORCES 


259 


Example.  A  weight  of  10  tons  is  raised  by  a  crane  whose  jib  is 
30'  long  and  stands  at  an  angle  of  45°  with  the  crane  post.  Find  the 
thrust  in  the  jib  if  the  tie-rod  is  attached  20' 
from  the  bottom  of  the  crane  post.  Since 
WCD  is  similar  to  ABC, 


C_— 


.,^^D 


5 


WD 
10 


30 
20 ' 


^10  Tona 


Fig.  182 


Thus  WD  =  15  tons,  the  thrust  in  the  jib. 
To  compute  CD  we  must  know  BC.     This  is 
found  from  the  triangle  ABC  as  follows  :  Draw  BK  perpendicular  to  AC. 
Then  BK  =  20  sin  45°  =  14.1,  AK  =  20  cos  45°  =  14.1, 

KC  =  30 -14.1  =  15.9. 
Therefore 

BC  =  V(14.1)2  +  (15.9)2  =  21.3 
Since 

CD  _BC  ^21.3 
10       20        20  ' 
it  follows  that 

CD  =  10.6  tons,  the  pull  in  the  tie-rod. 

When  the  weight  is  supported  by  a  rope  passed  over  pulleys  at  C 
and  B  and  on  to  a  drum  on  the  crane  post,  the  tension  in  the  tie-rod  is 
made  less  by  an  amount  equal  to  the  tension  in 
the  rope  which  supports  the  weight.  This  ten- 
sion is  the  weight  itself. 

Graphically,  if  CW  represents  the  weight 
and  therefore  the  tension  in  its  rope,  CD  is 
the  whole  tension  in  both  tie-rod  and  rope. 
Mark  off  CH  on  CD  and  equal  to  CW.  Then 
HD  must  be  the  tension  in  the  tie-rod  in  this  case. 


EXERCISES 

1.  A  weight  of  12  tons  is  suspended  rigidly  at  the  end  of  a  jib 
20  feet  long  by  a  tie-rod  15  feet  long.  Find  the  thrust  in  the  jib  and 
the  tension  in  the  tie-rod  if  the  tie-rod  is  attached  to  the  crane  post. 

(a)  10  feet  from  its  lower  end.  Ans.     24  tons ;  18  tons. 

(b)  13  feet  from  its  lower  end. 

(c)  15  feet  from  its  lower  end. 

(d)  20  feet  from  its  lower  end. 


260 


MATHEMATICS 


[XV,  §  219 


2.  In  Ex.  1,  find,  for  each  of  the  different  cases,  the  tension  in  the 
tie-rod  when  the  weight  is  supported  by  a  rope  which  passes  over  a 
pulley  at  C  and  is  led  parallel  to  the  tie-rod  over  a  pulley  at  B  and  on  to 
a  drum  on  the  crane  post. 

What  happens  in  cases  (c)  and  {d)  ? 

3.  If  AB  =  12',  AC  =  8',  and  TF  =  5  tons,  supported  rigidly  at  C, 
Fig.  181,  find  the  thrust  in  the  jib  and  the  tension  in  the  tie-rod  when 
BC  =  (a)  8  feet;  (6)  10  feet;  (c)  12  feet;  (d)   14  feet. 

4.  Discuss  the  previous  exercise  when  the  weight  is  supported  by 
a  rope  led  over  a  pulley  at  C,  parallel  to  the  tie-rod,  and  over  a  pulley 
atB. 

5.  The  load  suspended  rigidly  from  the  end  of  a  jib  is  5  tons.  The 
compression  in  the  jib  is  12  tons  and  the  inclination  of  the  jib  to  the 
horizontal  is  60°.     Find  the  tension  in  the  tie-rod.     (-,, 

6.  Discuss  Ex.  5  when  the  jib  is  inclined  to  the 
horizontal  at  an  angle  of  (a)  45°;  (b)  30°. 

7.  A  rod  AB,  30"  long,  hinged  at  A  and  carry- 
ing a  weight  of  150  lb.  at  B,  is  held  horizontally  by  a 
rope  BC  making  an  angle  of  45°  with  AB.  Find 
the  tension  in  the  rope. 

Let  the  segment  BD  =  150  lb.  in  the  triangle  BDE  similar  to  ABC, 
in  the  accompanying  figure.  Then  BE  on  the  same  scale  is  the  tension 
in  BC.     What  is  DE  ?  Ans.     Tension  =  150  V2  lb. 

8.  Solve  Ex.  7  if  BC  makes  an  angle  of  135° 
with  AB.     Also  if  it  makes  an  angle  of  120°  with  it. 

[Hint.  Let  CB  represent  150  lb.    To  find  x -50.] 

9.  A  weight  of  24  lb.  is  suspended  by  two 
flexible  strings  one  of  which  is  horizontal  and  the 
other  is  inclined  at  an  angle  of  30°  with  the  vertical. 
What  is  the  tension  in  each  string  ? 


lb.  \ 
Fig.  184 


-^E 


Fig.  185 


Ans.     8V3  1b.;  I6V3  lb. 


10.  Hay  is  pulled  up  at  the  end  of  a  bam  into  a  mow  by  fork,  pulleys, 
etc.  The  top  pulley  is  attached  to  a  beam  AB  fastened  at  A  to  the 
end  of  the  barn.  The  beam  is  held  horizontal  by  a  rope  attached  at 
B  and  making  an  angle  of  45°  with  AB.  What  is  the  tension  in  the 
rope  if  the  top  pulley  is  fastened  to  the  beam  4'  from  A  and  if  AB  is 
6'  long,  considering  a  load  of  300  lb.  on  the  pulley? 


XV,  §  220]  COMPOSITION  OF  FORCES  261 

Find  also  the  thrust  in  the  beam  against  the  end  of  the  barn. 

[Hint.  Determine  first  what  weight  at  B  will  produce  the  same 
moment  as  300  lb.  at  the  pulley.] 

11.  A  rod  10"  long,  hinged  at  one  end,  supports  a  weight  of  4  lb. 
S"  from  the  attached  end.  The  other  end  is  free  to  move  but  is  held 
horizontal  by  a  string  attached  to  this  end  and  making  an  angle  of  120® 
with  the  rod.     Find  the  tension  in  the  string.     Also  the  pull  in  the  rod. 

[Hint.     Find  first  the  equivalent  weight  at  the  end  of  the  rod.] 
Solve  also  when  the  weight  of  4  lb.  is  attached  to  the  rod  6"  from  the 
hinged  end. 

12.  A  man  weighing  200  lb.  stands  at  the  apex  of  a  roof  whose  equal 
rafters  are  10  ft.  long  on  a  building  16  ft.  wide.  Find  the  weight  tend- 
ing to  spread  the  walls  and  the  thrust  in  the  rafters. 

What  would  be  the  effect  of  shortening  the  rafters  for  the  same  width 
of  the  building  ? 

13.  When  a  roof,  as  the  one  in  Ex.  12,  is  covered  uniformly  with 
snow,  what  part  of  the  weight  of  the  snow  tends  to  spread  the  walls  ? 

220.  Resultant  of  Several  Forces  in  a  Plane  Acting  at  a 
Point. 

First  Method.  The  resultant  may  be  found  by  repeated 
application  of  the  triangle  of  forces.  Let  T^i,  F2,  F3,  F4  be 
forces  acting  upon  a  particle 
at  0.  The  resultant  of  Fi 
and  F2  is  OB,  Fig.  18G ;  of 
OB  and  F3  is  OC ;  of  OC 
and  F^isOD.  This  process 
may  be  continued  for  any 
number  of  forces  acting  in 
the  plane  at  a  point.  Thus 
OD  is  in  magnitude  and  direction  the  resultant  of  the  given 
forces. 

But  this  amounts  to  omitting  the  dotted  lines  and  draw- 
ing AB,  in  sense  and  direction,  equal  to  F2,  BC  similarly 


262 


MATHEMATICS 


CXV,  §  220 


equal  to  F3,  and  CD  similarly  equal  to  F4,  and  connecting  D 
to  0.  This  gives  rise  to  a  closed  polygon  called  the  polygon 
of  forces. 

The  polygon  of  forces  perhaps  reminds  one  of  the  dia- 
gram to  scale  of  a  closed  survey  in  which  the  last  course 
which  closes  the  survey  would  be  the  resultant.  This  idea 
gives  rise  to  a  second  method  of  finding  the  resultant. 

Second  Method.  Let  forces  Fi,  F2,  F3,  F4,  etc.,  in  the 
same  plane  act  on  a  particle  at  0.  Refer  the  forces  to  a 
pair  of  coordinate  axes  OX  and  OY.  In  surveying,  these  axes 
are  the  north-south'  line  and  the 
east- west  line.  Resolve  each  force 
into  its  rectangular  components 
with  reference  to  the  axes  as  shown 
in  Fig.  187.     For  example, 

OA  =  Fi  cos  «!, 

OB  =  Fi  sin  ai, 

etc.,  for  each  of  the  other  forces.  ^^^-  ^^^ 

//  a  component  acts  upward  or  to  the  right  it  is  positive; 
if  downward  or  to  the  left  it  is  negative.  For  example,  OD 
and  OG  are  negative. 

Let  us  denote  by  X  the  algebraic  sum  of  the  a:-components, 
and  by  Y  the  sum  of  the  ^/-components.  The  system  of 
forces  is  now  reduced  to  a  system  of  two  forces  at  right 
angles  to  each  other,  whose  resultant  has  a  magnitude 

(8)  i?  =  \/r+r. 

The  direction  of  the  resultant,  a,  is  given  by 


(9) 


Y 

tan  a=  — 

X 


where  a  is  the  angle  between  the  positive  direction   of  R 
and  the  positive  direction  of  the  a;-axis.     If  X  and  Y  are 


XV,  §  220]         COMPOSITION  OF  FORCES 


263 


plotted  properly,  the  quadrant  in  which 
a  lies  will  be  definitely  determined. 

Example.  Forces  of  6,  8,  10  lb.  act  on  a 
particle  at  angles  of  30°,  120°,  225°,  respectively, 
with  the  horizontal.  Find  the  magnitude  and 
direction  of  the  resultant. 

The  details  for  taking  the  rectangular  com- 
ponents may  be  arranged  as  follows. 


oTTa*'^ 


-7.1 


Fig.  188 


x-comp. 

j/-comp. 

Forces 

+ 

- 

+ 

- 

6 

5.2 

3 

8 

4 

6.9 

10 

7.1 

7.1 

Sums 

5.2 

11.1 

9.9 

7.1 

Hence  X=-5.9,     7  =+2.8 

Lay  off  on  the  diagram  X = OL  =  -  5.9  and  F  =  L72  =  2.8.     Then 


and 


OR 

tana  = 


V2:8'+5J9'  =  6.5, 
2.8 


-5.9 


=  -.4746 


Look  up  in  the  table  tan  a'  =  .4746  and  subtract  from  180®. 
Thus,  a'  =  25°  23'  with  the  negative  end  of  a:-axis  or 

a  =  154°  37',  the  direction  of  R  with  the  positive  x-axis. 

EXERCISES 

1.  Find  graphically  the  resultant  of  the  forces  in  the  example  of 
§  220  by  using  the  polygon  of  forces. 

2.  Forces  of  2  lb.,  3  lb.,  4  lb.  act  along  the  sides  of  an  equilateral 
triangle  taken  in  order.  Find  the  magnitude  wid  the  direction  and 
the  line  of  action  of  the  resultant. 

Find  the  resultants  of  the  following  systems  of  forces.  Check 
graphically. 

3.  100  lb.  acting  at  an  angle  of  0°  with  the  horizontal,  i.e.  (100  lb.,  0°), 
and  similarly  (50  lb.,  60°),  (200  lb.,  180°).         Ans.     (86.6  lb.,  150°). 


264 


MATHEMATICS 


PCV,  §  220 


4.  (4  lb.,  60°),  (5  lb.,  135°),  (6  lb.,  210°),  (8  lb.,  330°). 

5.  (9  lb.,  0°),  (7  lb.,  50°),  (5  lb.,  150°),  (3  lb.,  225°). 

6.  (250  lb.,  0°),  (270  lb.,  33°),  (160  lb.,  114°). 

Ans.     (502  lb.,  55°). 

7.  (270  lb.,  0°),  (132  lb.,  54°),  (215  lb.,  82°),  (138  lb.,  330°). 

Ans.     (558  lb.,  27°). 

8.  (35  lb.,  0°),  (24  lb.,  132°),  (54  lb.,  239°). 

Ans.     (297  lb.,  253°). 

9.  (60  lb.,  0°),  (58  lb.,  90°),  (86  lb.,  292°),  (110  lb.,  189°). 

Ans.     (42.3  lb.,  247°). 

221.  Resultant  of  Two  Parallel  Forces.  In  Fig.  189,  let 
APy  BQ  represent  two  parallel  forces  P  and  Q  acting  at  A 
and  B  respectively  upon  a  body 
in  the  same  direction.  The 
resultant  of  P  and  Q  is  desired. 

At  A  and  B  introduce  A  N 
=  BN',  any  two  equal  but  op- 
posite forces.  Since  their  re- 
sultant is  zero,  the  resultant  of 
P  and  Q  is  in  no  way  disturbed. 

At  A  draw  AP'  the  resultant  oi  AP  and  A  N  and  similarly 
at  B.     Produce  P'A  and  Q^B  to  meet  at  C. 

At  C  construct  the  parallelograms  of  forces  exactly  equal 
to  those  at  A  and  B  respectively.  Now  CN"  =  CN'"  and 
thus  neutralize  each  other  while  the  components  CP'",  and 
CQ"\  equal  respectively  to  P  and  Q,  act  in  the  line  CK 
parallel  to  the  lines  of  action  at  A  and  B.  Thus,  the  re- 
sultant of  P  and  Q  is  P+  Q. 


^11       c  N'" 

Fig.  189 


Moreover,  from  similar  triangles, 


AP 

AN 


CK 
AK' 


and 


BQ  _  CK 
BN'     BK' 


XV,  §  221]  COMPOSITION  OF  FORCES  265 

Dividing  these  two  equations,  member  for  member,  and 
remembering  that  AN=BN',  we  have 

AP^BK^b 
BQ      AK     a 

Therefore,  since  AP  represents  the  force  P,  and  BQ  repre- 
sents the  force  Q,  it  follows  that 

(10)  ^  =  -,    i.e.     Pa^Qh. 

Q     a 

A  similar  proof  may  be  given  for  the  case  of  unequal 
parallel  forces  acting  in  opposite  directions.  See  also  §§ 
198,  214.     Both  results  may  be  combined  as  follows. 

The  resultant  of  two  parallel  forces,  acting  in  the  same  or 
opposite  directions  at  the  extremities  of  a  rigid  straight  line,  is 
parallel  to  the  given  forces,  equal  to  their  algebraic  sum,  and 
divides  the  given  line  into  segments  which  are  inversely  pro- 
portional  to  the  given  forces. 

EXERCISES 

1.  A  lever  10  ft.  long  is  used  to  raise  a  weight  of  300  lb.  Where 
should  the  fulcrum  be  placed  if  a  man's  weight  of  150  lb.  is  available  at 
the  other  end  of  the  lever? 

2.  The  draft  of  a  plow  is  365  lb.  It  is  drawn  by  two  horses  hitched 
to  a  doubletree  of  which  the  end  clevis  holes  are  15  in.  and  19  in.,  re- 
spectively, from  the  middle  hole.     How  much  does  each  horse  pull? 

3.  Solve  Ex.  2  for  a  draft  of  p  lb.  and  end  clevis  holes  a  in.  and  h  in. 
from  the  middle  hole. 

4.  A  horse  weighs  1356  lb.  of  which  795  lb.  are  on  the  front  feet  and 
561  lb.  on  the  hind  feet.  If  the  distance  between  the  front  and  hind 
feet,  as  measured  at  their  points  of  contact  with  the  ground,  is  45  in., 
where  does  the  line  of  gravitation  fall  between  the  feet,  i.e.  where  is 
the  line  of  action  of  the  resultant  of  these  two  parallel  forces? 

5.  A  horse  weighing  1250  lb.  sustains  700  lb.  on  his  front  feet. 
The  distance  between  fore  and  hind  feet  is  48  in.  Where  is  the  line 
of  gravitation  ? 


266  MATHEMATICS  [XV,  §  221 

6.  If  the  line  of  gravitation  for  the  horse  in  the  previous  example 
falls  25  in.  to  the  rear  of  the  front  feet,  what  is  the  distance  between 
front  and  rear  feet? 

7.  Let  six  parallel  forces  proportional  to  the  numbers  1,  2,  3,  4,  5, 
6  act  at  the  points  (-2,  0),  (-1,  0),  (0,  0),  (1,  0),  (4,  0),  (-3,  0),  re- 
spectively.    Find  the  resultant  and  its  point  of  application. 

8.  On  a  straight  rod  AP  there  are  suspended  5  weights  of  5,  15, 
7,  6,  9  lb.  at  the  points  A,  B,  C,  D,  P,  respectively,  for  which  AB  =  S  ft., 
BC  =  6  ft.,  CD  =  5  ft.,  DP  =  4  ft.  Find  the  resultant  and  its  point  of 
application. 

9.  A  scantling  4"  x4"  Xl6'  is  supported  1  ft.  from  its  end.  What 
weight  will  it  lift  at  the  short  end  due  to  its  own  weight  when  horizontal 
if  it  weighs  4  lb.  per  linear  foot?  Solve  also  when  the  scantling  stands 
at  an  angle  of  30°  with  the  horizontal. 

[Hint.     The  weight  of  the  scantling  acts  at  its  center.] 
10.   A  bar  of  uniform  thickness,  10  ft.  long  and  weighing  150  lb., 
is  supported  at  its  extremities  in  a  horizontal  position,  and  carries  a 
weight  of  400  lb.  at  a  point  distant  3  ft.  from  one      ^  j^ 
extremity.     Find  the  pressures  on  the  points  of     "f 

support,  i.e.  at  A  and  at  D.  ^ — ^ — 2 P 

[Hint.     Let  x  =  pressure  at  A,  i.e.  x  lb.  would 


150  i6. 

just  raise  the  lever  AD.     Taking  moments  about  4oo^z6. 

D,  we  have  Fio.  190 

x(10)  ft.  =  (400)7  ft. +  (150)5  ft., 

i.e.  x  =  355  lb.  pressure  at  A.] 

11.  A  bridge  is  30  ft.  long  and  weighs  15  tons.  Find  the  pressure 
on  the  abutments  when  a  wagon  weighing  2  tons  is  one  third  of  the  way 
across.  Ans.     8^  tons ;  8f  tons. 

12.  The  wheels  of  a  wagon  oarry  a  load  of  800  lb.     What  extra 
horizontal  pull  is  necessary  to  draw  the  load  over 
an  obstruction  2"  high  when  the  wheels  are 

(a)  48"  in  diameter, 
(6)  42"  in  diameter, 
(c)  36"  in  diameter, 
{d)  30"  in  diameter? 

[Hint.  The  weight  acts  in  the  direction  OB.  Take  moments  about 
C,  i.e.  P  XDC  =  Weight  XAC]  Ans.     (a)  349  lb. 


XV,  §  221]         COMPOSITION   OF  FORCES 


267 


13.  A  carriage  wheel  whose  weight  is  W  and  radius  r  rests  upon  a 
level  road.  Show  that  the  pull  P  necessary  to  draw  the  wheel  over 
an  obstacle,  of  height  h,  is 


14.  What  height  of  obstacle  may  be  cleared  by  the  wheels  of  a 
wagon  carrying  a  load  of  800  lb.  if  the  wheels  are  36"  high  and  an  extra 
pull  of  150  lb.  is  available?  Ans.     .31" 

15.  Two  horses  weighing,  respectively,  1100  lb.  and  1500  lb.  are 
hitched  to  a  doubletree  38"  long.  Where  should  the  clevis  hole  be 
located  so  that  each  horse  shall  pull  in  proportion  to  his  weight? 

16.  Figure  192  represents  a  doubletree  with  end  holes  A  and  B 
placed  36  inches  apart  and  the  center  hole,  C,  3  inches  in  front  of  the 
mid-point  of  AB.  If  the  total  draft  of  the 
load  is  400  lb.,  what  is  each  horse  pulling 
when  the  doubletree  is  at  an  angle  of  60" 
with  the  line  of  draft? 

[Hint.   Compute  A //, //K,  and  X^.  Then 
Lx  times  AH  must  equal  L2  times  HE.   Why  ?] 

17.  Solve  Ex.  16  when  the  angle  made  by 
the  doubletree  with  the  line  of  draft  is  64°  33'. 

18.  Solve  Ex.  16,  if  the  middle  hole  C  is  4"  in  front  of  the  center  of  AB. 

19.  Where  should  C  be  located  so  that  the  forces  Li  and  L2  are 
equal  for  any  angle  the  evener  makes  with  the  line  of  draft?  Where 
should  it  be  so  that  the  horse  ahead  pulls  the  greater  load? 


APPENDIX 


TABLE  I.     Drainage.     A  fall  of  6"  to  T  for  each  100'  is  considered 
a  good  grade  for  farm  drainage. 

Let  A  =the  number  of  acres  to  be  drained. 

d  =  depth  in  inches  of  water  to  be  drained  off  in  24  hr. 
<  =  diameter  of  tile  in  inches. 
D  =  number  of  cubic  feet  of  water  discharged  per  second  for  a 

1  %  grade. 
X  =  the  fall  in  inches  per  100'. 

The  following  table  gives  D  for  tiles  of  different  sizes. 


Tile  diameter,  t .     . 

4 

6 

8 

10 

12 

15 

20 

Discharge,  D      .     . 

0.16 

0.49 

1.11 

2.05 

3.40 

6.29 

13.85 

Then 


A=^DVi2x. 
a 


TABLE  n.     Weights  of  Produce. 

(The  figures  represent  minimum  weights  per  bushel.) 

Barley 48  lb.  (in  Ore.,  46  lb. ;  Ala.,  Ga.,  Ky.,  Pa., 

47  lb. ;  Cal.,  50  lb. ;  La.,  32  lb.) 

Beans  (white)    .     .     .     .     60  lb. 

Bran 201b. 

Buckwheat 48  lb.  (in  Cal.,  40  lb. ;  Ky.,  56  lb. ;  Ida., 

N.  D.,  Okl.,  Ore.,  S.  D.,  Texas,  Wash., 
42  lb. ;  Kan.,  Minn.,  N.  C,  N.  J.,  O., 
Tenn.,  50  lb.) 

Clover  Seed 60  lb.  (in  N.  J.,  64  lb.) 

Corn  (ear) 70  lb.,  except  in  Miss.,  72  lb. ;    in  Ohio, 

68  lb. ;  in  Ind.  after  Dec.  1  and  in  Ky. 
after  May  1,  following  the  time  of  husk- 
ing, it  is  68  lb. 


270  MATHEMATICS 

Corn  (shelled)    ....  56  lb.  (in  Cal.,  52  lb.) 

Corn  (Meal)       ....  50  lb.  (in  Ala.,  Ark.,  Ga.,  lU.,  Miss.,  N.  C, 

Tenn.,  48  lb.) 

Flaxseed 56  1b. 

Grass  Seed  (blue)  ...  44  lb. 

Grass  Seed  (Hungarian)  50  lb. 

Hemp  Seed 44  lb. 

Millet  Seed 50  1b. 

Oats 32  lb.  (in  Ida.,  Ore.,  36  lb. ;  in  Md.,  26  lb;. 

in  N.  J.,  Va.,  30  lb.) 

Onions 57  lb. 

Peas 601b. 

Potatoes  (white)     ...  60  lb.  (in  Md.,  Pa.,  Va.,  56  lb.) 

Potatoes  (sweet)     .     .     .  55  lb. 

Rye 56  lb.  (in  Cal.,  54  lb. ;  in  La.,  32  lb.) 

Timothy  Seed    ....  45  lb.  (in  Ark.,  60  lb. ;  in  N.  D.,  42  lb.) 

Turnips 55  lb. 

Wheat 60  1b. 

TABLE  in.  Pounds  of  Nitrogen;  Phosphoric  Acid;  and  Potash 
required  per  acre,  as  a  minimum,  to  raise  the  following  crops.  Double 
the  figures  for  maximum. 

Cereal  Crops.  Barley:  12;  20;  25.  Buckwheat:  15;  30;  35. 
Corn  and  sorghum  :  10 ;  35 ;  30.  Oats  and  rye  :  12 ;  20 ;  30.  Wheat : 
12;  20;  12. 

Garden  Crops.  Asparagus :  20 ;  30 ;  35.  Cabbage  and  cauliflower : 
40;  70;  90.  Celery:  40;  50;  65.  Cucumbers,  muskmelon,  pump- 
kin, squash,  watermelon :  30 ;  50 ;  65.  Egg  plant :  40 ;  50 ;  90. 
Lettuce:  40;  50;  75.  Onions:  45;  55;  80.  Radishes:  15;  35;  45. 
Spinach:  15;  55;  40.     Tomatoes:  25;  35;  40. 

Grasses.  Lawns:  20;  25;  30.  Meadow,  millet:  15;  30;  35. 
Pasture:  15;  30;  40. 

Legumes.     Alfalfa,  clover :  5 ;  30 ;  40.     Beans,  peas :  5 ;  30 ;  35. 

Orchard  Crops,  Small  Fruits.  Apples,  pears,  quinces :  8 ;  30 ;  50. 
Blackberries :  15 ;  30 ;  40.  Cherries,  plums  :  10 ;  35 ;  45.  Currants, 
gooseberries :  10 ;  25 ;  40.  Grapes  :  8 ;  30 ;  45.  Nursery  stock : 
10;  25;  30.     Peaches:   15;  40;  55.     Strawberries:  25;  55;  70. 

Roots,  Tubers.  Beets,  turnips  :  20;  25;  35.  Carrots:  15;  35;  40. 
Horse  radish :  15 ;  25 ;  35.    Parsnips :  20 ;  55 ;  50.    Potatoes :  30 ;  40 ;  65. 


APPENDIX  271 

TABLE  rV.  Percentage  of  Nitrogen :  Phosphoric  Acid :  Potash  in 
Common  Fertilizing  Materials. 

(Analyses  vary  considerably.  The  following  figures  represent  rather 
high  grade  material.) 

Nitrogenous  Materials.  Dried  blood,  13  :  0 :  0.  Sodium  nitrate, 
15  :  0 :  0.  Ammonium  sulphate,  20 :  0 :  0.  In  analysis  and  guarantees, 
ammonia  =  about  .82  nitrogen. 

Phosphatic  Materials.  Acid  phosphate  (superphosphate,  dissolved 
stone,  acid  rock),  0 :  16  :  0.  Basic  slag  (Thomas  slag,  iron  phosphate), 
0:15:0.  Floats  (calcium  phosphate,  phosphate  of  lime,  raw  phos- 
phate, raw  rock),  0 :  12 :  0.     Acid  bone  black,  0 :  15  :  0. 

Combined  Nitrogenous  and  Phosphatic  Materials.  Acid  bone, 
2 :  20  :  0.  Bone  meal,  3  :  24  :  0.  Fish  scrap  (fish  guano,  fish  tankage), 
8:7:0.  Ground  tankage  (meat  guano,  slaughterhouse  refuse),  7:9:0. 
Steamed  bone,  2  :  26  :  0.     Manure  cake,  4:4:0. 

Potash  Materials.  Muriate  of  potash  (potassium  chloride),  0 :  0 :  50. 
Kainit,  0 :  0 :  12.     Sulphate  of  potash,  0 :  0 :  45. 

Combined  Phosphatic  and  Potash  Materials.  Wood  ashes,  0:1:5. 
Dry  ferns,  0 :  .37 :  1.8. 

Combined  Nitrogenous,  Phosphatic,  and  Potash  Materials.  Corn 
stalks,  .5:. 33:  1.67.  Guano  (Peruvian),  5:18:3.  Cottonseed  meal 
(cottonseed  cake),  7:2.5:1.5.  Farmyard  manure,  .5:. 5:. 5.  Hen 
manure,  dry,  2:2:1.  Leaves,  .7 :  .15  :  .3.  Oat  straw,  .7 :  .2  :  1.1.  Pea 
straw,  1 :  .3  :  1.     Sheep  manure,  2 :  1.5  :  1.5. 

TABLE  V.    Units  and  Equivalents  in  Weights  and  Measures. 
1  acre 43,560  sq.  ft.  =  160  sq.  rd. 

=  10  sq.  ch.  =40.47  ares  (metric) 
1  are  (metric) 100  sq.  meters  =  .01  hectare  (ha.) 

=  119.6  sq.  yd.  =  .02471  (1/40)  acre 

1  bag  cement 94  cu.  ft.  =94  lb. 

1  barrel 31^  gal.  =4|  cu.  ft. 

1  barrel  cement 4  bags  =  376  lb.  =3.76  cu.  ft. 

1  barrel  flour 196  lb. 

1  barrel  refined  oil 42  gal. 

1  barrel  salt 200  lb. 

1  brick 2"  X4'''  X8" 

1  bushel  (stroked) 2150.4   cu.    in.  =  li    cu.   ft. 

=  approximately  35.24  hters 


272  MATHEMATICS 

1  bushel  (heaped) 2688  cu.  in.  to  2747  cu.  in. 

1  bushel  of  coal 80  lb, 

1  centimeter  (cm.) 3937  in.  =  10  miUimeters 

1  chain 66  ft.  =  100  links  (li.) 

1  cubit 18  inches 

1  cubic  centimeter  water      .     .  1  gram  (at  4°  C.) 

1  cu.  in 16.387  c.c. 

leu.  ft =1/27    cu.    yd.  =  .037    cu.    yd. 

=  .0283  cubic  meter 

=  4/5  bu.  =7^  gal. 

1  cu.  ft.  brickwork 18  to  22  bricks 

1  cu.  ft  cement 100  lb. 

1  cu.  ft.  water 62,43  lb.  =  1000  oz.  approximately 

degrees  Fahrenheit  (F°)  .     .     .  F°  =  fC°+32° 

degrees  Centigrade  (C°)        .     .  C°  =  (F°-32°)f 

1  dram  (avoirdupois)  .     .     .     .  1/16  oz. 

1  fathom 6  feet 

1  foot 12  inches  =  30.48  cm. 

1  foot  (board  measure)     ...  a  board  1  ft.  square  and  1  in.   (or 

less)  thick 

1  furlong 40  rods 

1  gallon 4  quarts  =  3.7854  liters  =  231  cu.  in. 

1  gallon  20%  cream    "...  8.4  lb. 

1  gallon  22%  cream    ....  8.339  lb.  =  1  gallon  water 

1  gallon  milk  (3^%)     ....  8.6  lb. 

1  gallon  skim  milk 8.65  lb. 

1  gallon  water 8,339  lb. 

1  grain  (avoirdupois) 065  gram 

1  gram 15.432  grains  =  ,03527  oz. 

1  gross       ........  12  dozen 

1  hand 4  inches 

1  hectare  (ha.) 100  ares  =  2.471  acres 

1  hogshead 2bbl.=63gal. 

1  inch 2,54  cm,  =25.4  mm. 

1  kilogram,  or  kilo  (kg.)        .     .  1000  g.  =2.20462  lb. 

1  kilometer  (km.) 1000  meters  =  .62  mi. 

1  knot 1   nautical  mile  per  hour 

=  1.15  mi.  per  hour 

1  labor  (la-bor') 1,000,000  sq.  varas  =  173  acres 


APPENDIX  273 

1  league 3  nautical  mi.  =3.46  mi. 

1  league  (Spanish  unit  of  area)     25  labors  =  6.8  sq.  mi. 

llink 7.92  in.  =  .66  ft. 

1  liter 1000    cc.=244    cu.    in. 

=  1.057  qt.  (liquid)  =  .908  qt.  (dry) 

1  meter 39.37  in.  =3.3  ft. 

.91  meter 1  yard 

1  mil 1/6400  of  a  circumference 

=  .001  radian,  approximately 
1  mile 5280    ft.  =8    furlongs  =  320    rd. 

=  1.6  kilometers 

1  nautical  mile 1.153  mile  =  l  minute  on  equator 

1  millimeter 039in.  =  .lcm. 

1  oz.  (avoirdupois)       .     .     .     .     28.35  grams 

1  oz.  (Troy) 31.1  grams 

1  pace 3  feet 

1  penny  (English) $.0203  (before  1914) 

1  perch  =  1  pole 1  rod  =  16^  feet 

1  perch  of  stonework  ....     240  cu.  ft. 

TT  (pi) 3.1416  (See  p.  28.) 

1  pole (see  "perch") 

1  pound  (avoir.) 7000  grains  =  16  oz.  =  . 4536  kilogram 

1  pound  English  money  ...     20  shimngs  =  $4,866  (before  1914) 

1  pound  (Troy) 5760  grains  =  12  oz. 

1  quart  (hquid) 95  liter  (.9464) 

1  quart  (dry) 1.1  liter 

1  radian 180° -h7r  =  57.3°  =  1000  mils 

1  rod =1  perch  =  1  pole  =  16^  feet 

1  shiUing  (EngHsh)      ....     $.243  (before  1914) 

1  square  foot 144  square  inches 

1  square  inch 6.5  square  centimeters 

1  ton 2000  lb. 

1  ton  of  hay 500  cu.  ft.   (varies  from  450  cu.  ft. 

to  550  cu.  ft.  depending  on  quality) 
1  ton  of  coal 25  bu.   heaped    (2688    cu.   in.) 

=  38f  cu.  ft. 

1  vara  (va'ra) 33^  in.  in  Texas 

33  in.  in  Cal.  and  Mexico 
1  yard 3  ft.  =  .91  meter 


274 


MATHEMATICS 


TABLE  VI.     American  Experience  MortaKty  Table. 


At 

Number 

At 

Number 

At 

Number 

At 

Number 

Age. 

Surviving. 

Deaths. 

Age. 

Surviving. 

Deaths. 

Age. 

Surviving. 

Deaths. 

Age. 

Surviving. 

Deaths. 

10 

100,000 

749 

35 

81,822 

732 

60 

57,917 

1,546 

85 

5,485 

1,292 

11 

99,251 

746 

36 

81,090 

737 

61 

56,371 

1,628 

86 

4,193 

1,114 

12 

98,505 

743 

37 

80,353 

742 

62 

54,743 

1,713 

87 

3,079 

933 

13 

97,762 

740 

38 

79,611 

749 

63 

53,030 

1,800 

88 

2,146 

744 

14 

97,022 

737 

39 

78,862 

756 

64 

51,230 

1,889 

89 

1,402 

555 

15 

96,285 

735 

40 

78,106 

765 

65 

49,341 

1,980 

90 

847 

385 

16 

95,550 

732 

41 

77,341 

774 

66 

47,361 

2,070 

91 

462 

246 

17 

94,818 

729 

42 

76,567 

785 

67 

45,291 

2,158 

92 

216 

137 

18 

94,089 

727 

43 

75,782 

797 

68 

43,133 

2,243 

93 

79 

58 

19 

93,362 

725 

44 

74,985 

812 

69 

40,890 

2,321 

94 

21 

18 

20 

92,637 

723 

45 

74,173 

828 

70 

38,569 

2,391 

95 

3 

3 

21 

91,914 

722 

46 

73,345 

848 

71 

36,178 

2,448 

22 

91,192 

721 

47 

72,497 

870 

72 

33,730 

2,487 

23 

90,471 

720 

48 

71,627 

896 

73 

31,243 

2.505 

24 

89,751 

719 

49 

70,731 

927 

74 

28,738 

2,501 

25 

89,032 

718 

50 

69,804 

962 

75 

26,237 

2,476 

26 

88,314 

718 

51 

68,842 

1,001 

76 

23,761 

2,431 

27 

87,596 

718 

52 

67,841 

1,044 

77 

21,330 

2,369 

28 

86,878 

718 

53 

66,797 

1,091 

78 

18,961 

2,291 

29 

86,160 

719 

54 

65,706 

1,143 

79 

16,670 

2,196 

30 

85,441 

720 

55 

64,563 

1,199 

80 

14,474 

2,091 

31 

84,721 

721 

56 

63,364 

1,260 

81 

12,383 

1,964 

32 

84,000 

723 

57 

62,104 

1,325 

82 

10,419 

1,816 

33 

83,277 

726 

58 

60,779 

1,394 

83 

8,603 

1,648 

34 

82.551 

729 

59 

59,385 

1,468 

84 

6,955 

1,470 

Certain  Convenient  Values  for  n  =  1  to  n=10 


n 

1/n 

Vn 

Vn       ) 

n! 

l/n\ 

LoGio  n 

1 

1.000000 

1.00000 

1.00000 

1 

1.0000000 

0.000000000 

2 

0.500000 

1.41421 

1.25992 

2 

0.5000000 

0.3010299f)6 

3 

0.333333 

1.73205 

1.44225 

6 

0.1G66667 

0.477121255 

4 

0.250000 

2.00000 

1.58740 

24 

0.0416667 

0.602059991 

5 

0.200000 

2.23607 

1.70998 

120 

0.0083333 

0.698970004 

6 

0.166667 

2.44949 

1.81712 

720 

0.0013889 

0.778151250 

7 

0.142857 

2.64575 

1.91293 

5040 

0.0001984 

0.845098040 

8 

0.125000 

2.82843 

2.00000 

40320 

0.0000248 

0.903089987 

9 

0.111111 

3.00000 

2.08008 

362880 

0.0000028 

0.954242509 

10 

0.100000 

3.16228 

2.15443 

3628800 

0.0000003 

1.000000000 

The  tables  on  the  following  pages  are  four-place  tables  of  logarithms, 
trigonometric  functions,  powers  and  roots,  and  compound  interest. 
More  extensive  tables  on  all  these  may  be  found  in  The  Macmillan 
Tables,  from  which  these  are  taken. 


TABLES 


275 


TABLE  VII.    Squares  and  Cubes,  Square  Roots,  and  Cube  Roots. 


No. 

Squabe 

Cubs 

Square 

KoOT 

Cube 
Root 

No. 

Square 

Cube 

Square 
Root 

Cube 
Root 

1 

1 

1 

1.000 

1.000 

61 

2,601 

132,651 

7.141 

3.708 

2 

4 

8 

1.414 

1.260 

52 

2,704 

140,608 

7.211 

3.733 

3 

9 

27 

1.732 

1.442 

53 

2,809 

148,877 

7.280 

3.750 

4 

16 

64 

2.000 

1.587 

54 

2,916 

157,464 

7.348 

3.780 

5 

25 

125 

2.236 

1.710 

55 

3,025 

1(KJ,375 

7.416 

3.803 

6 

36 

216 

2.449 

1.817 

56 

3,136 

175,616 

7.483 

3.826 

7 

49 

343 

2.646 

1.913 

57 

3,249 

185,193 

7.550 

3.849 

8 

64 

512 

2.828 

2.000 

58 

3,364 

195,112 

7.616 

3.871 

9 

81 

729 

3.000 

2.080 

59 

3,481 

205,379 

7.681 

3.893 

10 

100 

1,000 

3.162 

2.154 

60 

3,000 

216,000 

7.746 

3.915 

11 

121 

1,331 

3.317 

2.224 

61 

3,721 

226,981 

7.810 

3.936 

12 

144 

1,728 

3.464 

2.289 

62 

3,844 

238,328 

7.874 

3.958 

13 

169 

2,197 

3.606 

2.351 

63 

3,969 

250,047 

7.937 

3.979 

14 

196 

2,744 

3.742 

2.410 

64 

4,096 

262,144 

8.000 

4.000 

15 

225 

3,375 

3.873 

2.466 

65 

4,225 

274,625 

8.062 

4.021 

16 

256 

4,096 

4.000 

2.520 

66 

4,356 

287,496 

8.124 

4.041 

17 

289 

4,913 

4.123 

2.571 

67 

4,489 

300,763 

8.185 

4.062 

18 

324 

6,832 

4.243 

2.621 

68 

4,624 

314,432 

8.246 

4.082 

19 

361 

6,859 

4.359 

2.668 

69 

4,701 

328,509 

8.307 

4.102 

20 

400 

8,000 

4.472 

2.714 

70 

4,tX)0 

343,000 

8.367 

4.121 

21 

441 

9,261 

4.583 

2.759 

71 

5,041 

357,911 

8.426 

4.141 

22 

484 

10,648 

4.090 

2.802 

72 

5,184 

373,248 

8.485 

4.160 

23 

529 

12,167 

4.796 

2.844 

73 

5,329 

389,017 

8.544 

4.179 

24 

576 

13,8^4 

4.899 

2.884 

74 

5,476 

405,224 

8.602 

4.198 

25 

625 

15,625 

5.000 

2.924 

75 

5,625 

421,875 

8.660 

4.217 

26 

676 

17,576 

5.099 

2.962 

76 

6,776 

438,976 

8.718 

4.236 

27 

729 

19,683 

5.1% 

3.000 

77 

5,929 

456,533 

8.775 

4.254 

28 

784 

21,952 

5.292 

3.037 

78 

6,084 

474,552 

8.832 

4.273 

29 

841 

24,389 

5.385 

3.072 

79 

6,241 

493,039 

8.888 

4.291 

30 

900 

27,000 

5.477 

3.107 

80 

6,400 

512,000 

8.944 

4.309 

31 

961 

29,791 

5.568 

3.141 

81 

6,561 

531,441 

9.000 

4.327 

32 

1,024 

32,768 

5.657 

3.175 

82 

6,724 

551,368 

9.055 

4.344 

33 

1,089 

35,937 

5.745 

3.208 

83 

6,889 

571,787 

9.110 

4.362 

34 

1,156 

39,304 

6.831 

3.240 

84 

7,056 

592,704 

9.165 

4.380 

35 

1,225 

42,875 

5.916 

3.271 

85 

7,225 

614,125 

9.220 

4.397 

36 

1,296 

46,656 

6.000 

3.302 

86 

7,396 

636,056 

9.274 

4414 

37 

1,369 

50,653 

6.083 

3.332 

87 

7,569 

658,503 

9.327 

4.431 

38 

1,444 

54,872 

6.164 

3.362 

88 

7,744 

681,472 

9.381 

4.448 

39 

1,521 

59,319 

6.245 

3.391 

89 

7,921 

704,969 

9.434 

4.465 

40 

1,600 

64,000 

6.325 

3.420 

90 

8,100 

729,000 

9.487 

4.481 

41 

1,681 

68,921 

6.403 

3.448 

91 

8,281 

753,571 

9.539 

4.498 

42 

1,764 

74,088 

6.481 

3.476 

92 

8,464 

778,688 

9.592 

4.514 

43 

1,849 

79,507 

6.557 

3.503 

93 

8,649 

804,357 

9.644 

4.531 

44 

1,936 

85,184 

6.633 

3.530 

94 

8,836 

830,584 

9.695 

4.547 

45 

2,025 

91,125 

6.708 

3.557 

95 

9,025 

857,375 

9.747 

4.563 

46 

2,116 

97,336 

6.782 

3.583 

96 

9,216 

884,736 

9.798 

4.579 

47 

2,209 

103,823 

6.856 

3.609 

97 

9,409 

912,673 

9.849 

4.595 

48 

2,304 

110,592 

6.928 

3.634 

98 

9,604 

941,192 

9.899 

4.610 

49 

2,401 

117,649 

7.000 

3.659 

99 

9,801 

970,299 

9.950 

4.626 

50 

2,500 

125,000 

7.071 

3.684 

100 

10,000 

1,000,000 

10.000 

4.642 

276 

Table  VIII  — 

Four  Place  Logarithms 

N 

0 

1 

2 

3 

4|$ 

6 

7 

8 

9 

12  3 

4  5  6 

7  8  9 

10 

0000 

0043 

0086 

0128 

0170'  0212 

0253 

02f>4  0334 

0374 

4  8  12 

17  21  25 

29  33  37 

11 
12 
13 

14 
15 

16 

17 
18 
19 

0414 
0792 
1139 

1461 
1761 
2041 

2304 
2553 

2788 

0453 
0828 
1173 

1492 

1790 
2068 

2330 
2577 
2810 

0492 
0864 
1206 

1523 
1818 
2095 

2355 
2(501 
2833 

0531 
0899 
1239 

1553 
1847 
2122 

2380 
2625 
2856 

05(59 
0934 
1271 

1584 
1875 
2148 

2405 
2648 

2878 

0607 
0969 
1303 

1614 
1903 
2175 

2430 
2672 
2900 

0645 
1004 
1335 

1644 
1931 
2201 

2455 
2695 
2923 

0682 
1038 
1367 

1673 
1959 

2227 

2480 
2718 
2945 

0719 
1072 
1399 

1703 
1987 
2253 

2504 
2742 
29(57 

0755 
1106 
1430 

1732 
2014 
2279 

2529 
2765 
2989 

4  811 
3  7  10 
3  6  10 

3  6  9 
3  6  8 
3  5  8 

2  5  7 
2  5  7 
2  4  7 

15  19  23 
14  17  21 
13  16  19 

12  15  18 
11  14  17 
11  13  16 

10  12  15 
9  12  14 
9  11  13 

26  30  34 
24  28  31 
23  26  29 

21  24  27 
20  22  25 
18  21  24 

17  20  22 
16  19  21 
16  18  20 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

2  4  6 

8  11  13  1  15  17  19  1 

21 1 

23  j 

24  1 
25 
26 

27 
28 
29 

30 

31 
32 
33 

34 
35 

36 

37 
38 
39 

3222 
3424 
3617 

3802 
3979 
4150 

4314 
4472 
4624 

4771 

3243 
3444 
3636 

3820 
3997 
4166 

4330 
4487 
4(>3i) 

4786 

3263 
3464 
3655 

3838 
4014 
4183 

4346 

4502 
4(;54 

4800 

3284 
3483 
3674 

3a56 
4031 
4200 

4362 

4518 
4Wi9 

3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 

3324 
3522 
3711 

3892 
4065 
4232 

4393 
4548 
4698 

3345 
3.541 
3729 

3909 
4082 
4249 

4409 
45(54 
4713 

3.365 
35(50 
3747 

3927 
4099 
4265 

4425 
4579 

4728 

3385 
3.579 
3766 

3945 
4116 
4281 

4440 
4594 
4742 

3404 

3598 
3784 

3962 
4133 

4298 

4456 
4609 
4757 

2  4  6 
2  4  6 
2  4  6 

2  4  5 
2  4  5 
2  3  6 

2  3  5 
2  3  5 
13  4 

8  1012 
8  10  12 
7  9  11 

7  9  11 
7  9  10 
7  8  10 

6  8  9 
6  8  9 
6  7  9 

14  16  18 
14  16  17 
13  15  17 

12  14  16 
12  14  16 
11 13  15 

11 12  14 
11  12  14 
10  12  13 

4814 

4829 

4843 

4857 

4871 

4886 

4<)00 

13  4 

6  7  9  1  10  11 13  1 

4914 
5051 
5185 

5315 
5441 
5563 

5682 
5798 
5911 

4928 
50(>5 
5198 

5328 
5453 
5575 

5694 
58m) 
5922 

4942 
5079 
5211 

5340 
5465 

5587 

5705 
5821 
5933 

4955 
5092 
5224 

5353 
5478 
5599 

5717 
5832 
5944 

4969 
5105 
5237 

5366 
5490 
5611 

5729 
5843 
5955 

4983 
5119 
5250 

5378 
5502 
5623 

5740 

5855 
59f56 

4997 
5132 
5263 

5391 
5514 
5635 

5752 
5866 
5977 

5011 
5145 
5276 

5403 
5527 
5647 

5763 

5877 
5988 

5024 
5159 
5289 

5416 

5539 
5658 

5775 
5888 
599<) 

5038 
5172 
5302 

5428 
5.551 
5670 

5786 
5899 
6010 

1  3  4 
1  3  4 
13  4 

12  4 
12  4 
12  4 

12  4 

1  2  3 
1  2  3 

5  7  8 
5  7  8 
5  7  8 

5  6  8 
5  6  7 
5  6  7 

5  6  7 
5  6  7 

4  5  7 

10  11  12 
91112 
9  1112 

91011 
910  11 
81011 

8  911 
8  9  10 
8  910 

40 

6021 

0031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

12  3 

4  5  6 

8  9  10 

41 
42 
43 

44 
45 

46 

47 
48 
49 

6128 
6232 
63a^ 

6435 
6532 
6628 

(J721 
6812 
6902 

6138 
6243 
6345 

6444 
6542 
6637 

6730 
()821 
()911 

6149 
6253 
6355 

6454 
(;551 
6646 

6739 
6830 
6920 

7007 

6160 
6263 
6365 

6464 
65()1 
6656 

6749 
683i) 
6928 

7016 

6170 
6274 
6375 

6474 
6571 
6665 

6758 
(5848 
6937 

6180 
6284 
6385 

6484 
a580 
6675 

6767 
6857 
694(5 

6191 
6294 
6395 

6493 
6590 
(J084 

6776 
68(56 
6955 

6201 
6,304 
6405 

6503 
6599 
6693 

67a5 
6875 
69(W 

6212 
6314 
6415 

6513 
6609 
6702 

6794 

6884 
6972 

6222 
6325 
6425 

6522 
6618 
6712 

6803 
6893 
6981 

12  3 
12  3 
12  3 

12  3 
12  3 
12  3 

12  3 
12  3 
12  3 

4  5  6 
4  5  6 
4  6  6 

4  5  6 
4  5  6 
4  6  6 

4  5  6 
4  5  6 

4  4  5 

7  8  9 
7  8  9 
7  8  9 

7  8  9 
7  8  9 
7  7  8 

7  7  8 
7  7  8 
6  7  8 

50 

6990 

6998 

7024 

7033 

7042 

7050 

7059 

7067 

1  2  3 

3  4  6  1  6  7  8 

51 
52 
53 

54 

7076 
7160 
7243 

7324 

7084 
7168 
7251 

7332 

7093 

7177 
7259 

7340 

7101 
7185 
7267 

7348 

7110 
7193 
7275 

7.356 

7118 
7202 
7284 

7364 

7126 
7210 
7292 

7372 

7135 
7218 
7300 

7380 

7143 
7226 
7308 

7388 

71.52 
72.35 
7316 

73^)6 

12  3 
12  3 
12  2 

1  2  2 

3  4  6 
3  4  5 
3  4  6 

3  4  5 

6  7  8 
6  7  7 
6  6  7 

6  6  7 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

12  2 

4  5  6 

7  8  9 

The  proportional  parts  are  stated  in  full  for  every  tenth  at  the  right-hand  side. 
The  logarithm  of  any  number  of  four  significant  figures  can  be  read  directly  by  add- 


Table  Till  — Four  Place  Logarithms 


277 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

12    3 

4   5   6 

7    8    9 

55 

56 

57 
58 
59 

7404 
7482 

7559 
7634 

7709 

7412 
7490 

7566 
7642 
7716 

7419 
7497 

7574 
7649 
7723 

7427 
7505 

7582 
7657 
7731 

7435 
7513 

7589 

7738 

7443 
7520 

7597 
7(572 
7745 

7451 
7528 

7(504 
7679 
7752 

7459 
7536 

7612 
7(586 
7760 

7466 
7543 

7619 
7(5i)4 
7767 

7474 
7551 

7627 
7701 

7774 

1   2   2 
12    2 

1    1   2 
112 
112 

3   4    5 
3   4   5 

3   4    5 
3    4    4 
3   4   4 

5  6  7 
5   6   7 

5  6  7 
5  6  7 
5    6    7 

60 

61 
62 
63 

64 
65 

66 

67 
68 
69 

7782 

7789 

7860 
7931 
8000 

8069 
8136 
8202 

8267 
8331 
8395 

7796 

7803 

7810 

7882 
7952 
8021 

8089 

8222 

8287 
8;J51 

7818 

7825 

7832 

7839 

7846 

1    1   2 

3   4   4 

5   6   6 

7853 
7924 
7993 

8062 
8129 
8195 

8261 
8325 

8388 

7868 
7938 
8007 

8075 
8142 
8209 

8274 
83;38 
8401 

7875 
7945 
8014 

8082 
8149 
8215 

8280 
8:U4 
8407 

7889 
7959 
8028 

8096 
8162 

8228 

8293 
8357 
8420 

7896 
7;K56 
8035 

8102 
8169 
8235 

8299 
8363 
8426 

7903 
7973 
8041 

8109 
8176 
8241 

8306 
8370 
8432 

7910 

7980 
8048 

8116 

8182 
8248 

8312 

8376 
8439 

7917 
7987 
8055 

8122 

8189 
8254 

8319 

8:^2 

8445 

112 
112 
112 

1    1    2 
112 
112 

112 
112 
112 

3    3   4 
3    3    4 
3   3    4 

3    3   4 
3    3    4 
3    3   4 

3   3   4 
3    3    4 
3    3    4 

5  6  6 
5  5  6 
5   6   6 

5  5  6 
5  5  6 
5    5   6 

5  5  6 
4   5   6 

4    5    6 

70 

W51 

8457 

M(\3 

S470 

8476 

8482 

8488 

8494 

8500 

8.50(5 

112 

3   3    4 

4   5   6 

71 
72 
73 

74 
75 

76 

77 
78 
79 

8513 
8573 
8633 

8692 
8751 
8808 

8865 
8921 
8976 

8519 
8579 
8639 

8698 
8756 
8814 

8871 
8927 
8982 

8525 
8585 
8645 

8701 
87(52 
8820 

8876 
8932 
8987 

8531 
8591 
8651 

8710 

8768 
8825 

8882 
8938 
85)93 

8537 
8597 
86.57 

8716 
8774 
8831 

8887 
8943 
8<)98 

a543 
8(503 
8663 

8722 
8779 
8837 

8893 
81H!) 
9004 

a549 
8(509 
8(569 

8727 
8785 
8842 

88W 
8954 
9009 

a555 
8(515 
8675 

8733 
8791 
8848 

8904 
89(50 
<)015 

8561 
8(521 
8681 

8739 

8797 
8854 

8910 
8<Xv) 
9020 

a567 
8(527 
8686 

8745 
8802 
8859 

8915 
8971 
9025 

112 
112 
112 

112 
112 
112 

1    1    2 
112 
112 

3    3   4 
3    3    4 
2    3   4 

2    3   4 
2    3    3 
2    3    3 

2    3    3 
2    3   3 
2    3   3 

4  5  6 
4  5  6 
4   5    5 

4  5  5 
4  5  5 
4   4   5 

4  4  5 
4  4  5 
4    4   5 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

112 

2    3    3 

4    4   5 

81 
82 
83 

H4 
85 

86 

87 

88 
89 

908.5 
9138 
9191 

9243 
92<)4 
9345 

9395 
9445 
{U94 

9090 
9143 
9196 

9248 
9299 
9350 

9100 
9450 

9096 
9149 
9201 

9253 
9;?04 
9355 

9405 
9455 
9504 

9101 
9154 
9206 

9258 
9309 
93(50 

9410 

<H(i0 
950<) 

910(5 
9159 
9212 

9263 
9315 
9365 

9415 
94(55 
9513 

9112 
9165 
9217 

9269 
9320 
9370 

9420 
M69 
9518 

9117 
9170 
9222 

9274 
9325 
9375 

9425 
9474 
9523 

9122 
9175 
9227 

9279 

93:jo 

9380 

9430 
i)479 

9528 

9128 
9180 
9232 

9284 
9;535 
9385 

9435 
9484 
9533 

9133 
9186 
9*238 

9289 
9'MO 
9390 

9440 

9489 
9538 

112 
112 
112 

112 
112 
112 

112 
Oil 
0    1    1 

2    3    3 
2    3    3 
2    3   3 

2    3    3 
2    3    3 
2   3    3 

2    3    3 
2   2    3 
2    2    3 

4   4    5 

4  4  5 
4   4    5 

4  4  5 
4  4  5 
4   4   5 

4  4  5 
3  4  4 
3    4    4 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

Oil 

2    2    3 

3    4    4 

91 
92 
93 

94 
95 

96 

97 
98 
99 

9590 
%38 
9685 

9731 

9777 
9823 

9868 
9912 
9956 

9595 
9643 
9689 

9736 
9782 
9827 

9872 
9917 
9961 

9600 
9647 
9694 

9741 
9786 
9832 

9877 
9921 
99f)5 

9(i05 
9(i52 
9699 

9745 
9791 
9836 

9881 

9<:»2e; 

9969 

9609 
9657 
9703 

9750 
9795 
9841 

9886 
99;J0 
9974 

9614 
9661 
9708 

9754 
9800 
9845 

9890 
9931 
9978 

9619 
966(5 
9713 

9759 
9805 
9850 

9894 
9939 
9983 

%24 
9(571 
9717 

9763 
9809 
9854 

9899 
9<H3 
9987 

9628 
9675 
9722 

9768 
9814 
9859 

<)903 
9948 
9991 

1^)33 
9(580 
9727 

9773 

9818 
9803 

9908 
9952 
9996 

0    1    1 
Oil 
Oil 

0    1    1 
0    1    1 
0    1    1 

0    1    1 
Oil 
Oil 

2    2    3 
2    2    3 
2    2    3 

2    2    3 
2    2    3 
2    2    3 

2    2    3 

2    2    3 
2    2    3 

3  4  4 
3  4  4 
3    4    4 

3  4  4 
3  4  4 
3   4    4 

3  I  4 
3  3  4 
3    3    4 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

12    3 

4    5    6 

7    8    9 

ing  the  proportional  part  corresponding  to  the  fourth  figure  to  the  tabular  number 
corresponding  to  the  first  three  figures.     There  may  be  an  error  of  1  in  the  last  place. 


278 


Table  IX  — Four  Place  Trigonometric  Tunctions 


[Characteristics  of  Log'arithina  om 

tted- 

determir 

le  by  the  usual  rule  from  the  val 

ue] 

Radians 

Degrees 

Sine 

Tangent 

Cotangent 

Cosine 

\^alue 

Lo?io 

Value 

Logio 

Value 

Logio 

Value   Logio 

.0000 

0°00' 

.0000 

.(XX)0 

1.0000  .0000 
1.0000  .0000 

90°  00' 

50 

1.5708 
1.5679 

!0029 

10 

.0029 

.4637 

.0029 

.4637 

343.77 

.5363 

.0058 

20 

.00.18 

.7(548 

.0058 

.7648 

171.89 

.2352 

1.0000  .0000 

40 

1.5650 

.0087 

30 

.0087 

.9408 

.0087 

.9409 

114.59 

.0591 

1.0000  .0000 

30 

1.5621 

.0116 

40 

.0116 

.0(558 

.0116 

.0(558 

85.940 

.9342 

.9999  .0000 

20 

1.5592 

.0145 

50 

.0145 

.1(527 

.0145 

.1627 

68.750 

.8373 

.9999  .0000 

10 

1.5563 

.0175 

1°00' 

.0175 

.2419 

.0175 

.2419 

57.290 

.7581 

.9998  .9999 

89°  00' 

1.5533 

.0204 

10 

.0204 

.3088 

.0204 

.3089 

49  104 

.6911 

.9998  .9999 

50 

1.5504 

.0233 

20 

.0233 

.3608 

.0233 

.36(59 

42.964 

.(5331 

.9997  .9999 

40 

1.5475 

.0262 

30 

.0262 

.4179 

.02(52 

.4181 

38.188 

.5819 

.9997  .9m) 

30 

1.5446 

.0291 

40 

.0291 

.4637 

.0291 

.4638 

34.368 

.5362 

.9i)96  .9998 

20 

1.5417 

.0320 

50 

.0320 

.5050 

.0320 

.5053 

31.242 

.4947 

.9995  .9998 

10 

1.5388 

.0349 

2°  00' 

.0349 

.5428 

.0349 

.5431 

28.636 

.4569 

.9994  .9997 

88°  00' 

1.5359 

.0378 

10 

.0378 

.5776 

.0378 

.5779 

2(5.432 

.4221 

.9993  .9997 

50 

1.5330 

.0407 

20 

.0107 

.6097 

.0407 

.6101 

24.542 

.3899 

.9992  .9<>96 

40 

1.5301 

.0436 

30 

.04^6 

.6397 

.0437 

.6401 

22.iX)4 

.3599 

.99f»0  .9996 

30 

1.5272 

.0465 

40 

.04(55 

.6677 

.0466 

.6(582 

21.470 

.3318 

.9989  .9995 

20 

1.5243 

.0495 

50 

.0494 

.6940 

.0495 

.6945 

20.206 

.3055 

.9988  .9995 

10 

1.5213 

.0524 

3°  00' 

.0523 

.7188 

.0524 

.7194 

19.081 

.2806 

.9986  .9994 

87°  00' 

1.5184 

.0553 

10 

.0552 

.7423 

.0553 

,7429 

18.075 

.2571 

.9985  .i]993 

50 

1.5155 

.0582 

20 

.0581 

.7645 

.0582 

.7652 

17.169 

.2348 

.9983  .9993 

40 

1.5126 

.0611 

30 

.0610 

.7857 

.0612 

.78(55 

16..350 

.2135 

.9981  .9992 

30 

1.5097 

.0640 

40 

.0640 

.8059 

.0(541 

.80(57 

15.605 

.1933 

.9980  .9991 

20 

1.5068 

.0669 

50 

.0669 

.8251 

.0070 

.8261 

14.924 

.1739 

.9978  .9990 

10 

1.5039 

.0698 

4°  00' 

.0(598 

.843(3 

.0699 

.8446 

14.301 

.1554 

.9976  .9989 

86° 00' 

1.5010 

.0727 

10 

.0727 

.8613 

.0729 

.8(524 

13.727 

.1376 

.9974  .9989 

50 

1.4981 

.0756 

20 

.0756 

.8783 

.0758 

.8795 

13.197 

.1205 

.9971  .9988 

40 

1.4952 

.0785 

30 

.0785 

.8946 

.0787 

.89(50 

12.706 

.1040 

.9969  .9987 

30 

1.4923 

.0814 

40 

.0814 

.9104 

.0816 

.9118 

12.251 

.0882 

.9967  .9986 

20 

1.4893 

.0844 

50 

.0843 

.9256 

.0846 

.9272 

11.826 

.0728 

.9964.  .9985 

10 

1.4864 

.0873 

5°  00' 

.0872 

.9403 

.0875 

.9420 

11.430 

.0580 

.9962  .9f)83 

85°  00' 

1.4835 

.0902 

10 

.0901 

.9545 

.0^)04 

.9563 

11.059 

.0437 

.9959  .9982 

50 

1.4806 

.0931 

20 

.0929 

.%82 

.0934 

.9701 

10.712 

.0299 

.9957  .9981 

40 

1.4777 

.09()0 

30 

.0958 

.9816 

.0%3 

.9836 

10.385 

.0164 

.9954  .9980 

30 

1.4748 

.0989 

40 

.0987 

.9945 

.0^)92 

.«)66 

10.078 

.0034 

.9951  .9979 

20 

1.4719 

.1018 

50 

.1016 

.0070 

.1022 

.0093 

9.7882 

.9907 

.9948  .9977 

10 

1.4690 

.1047 

6°  00' 

.1045 

.0192 

.1051 

.0216 

9.5144 

.9784 

.9945  .9976 

84°  00' 

1.4661 

.1076 

10 

.1074 

.0311 

.1080 

.0336 

9.2553 

.9(564 

.9942  .9975 

50 

1.4632 

.1105 

20 

.1103 

.0426 

.1110 

.0453 

9.0(^98 

.9547 

.9939  .9973 

40 

1.4603 

.1134 

30 

.1132 

.0539 

.1139 

.0567 

8.7769 

.9433 

.9936  .9972 

30 

1.4573 

.1164 

40 

.1161 

.0(>48 

.11(59 

.0(578 

8.5555 

.9322 

.t)932  .9ii71 

20 

1 .4544 

.1193 

50 

.1190 

.0755 

.1198 

.0786 

8.3450 

.9214 

.9929  .9969 

10 

1.4515 

.1222 

7°  00' 

.1219 

.0859 

.1228 

.0891 

8.1443 

.9109 

.9925  .9968 

83°  00' 

1.4486 

.1251 

10 

.1248 

.0i>61 

.1257 

.0995 

7.9530 

.m)5 

.9922  .99(5(5 

50 

1.4457 

.1280 

20 

.1276 

.10(50 

.1287 

.1096 

7.7704 

.8904 

.9918  .9964 

40 

1.4428 

.1309 

30 

.1305 

.1157 

.1317 

.1194 

7.5958 

.8806 

.9i)14  .9963 

30 

1.4399 

.1338 

40 

.1334 

.1252 

.1346 

.1291 

7.4287 

.8709 

.9911  .9961 

20 

1.4370 

.1367 

50 

.1363 

.1345 

.1376 

.1385 

7.2687 

.8615 

.9907  .9959 

10 

1.4341 

.1396 

8°  00' 

.1392 

.1436 

.1405 

.1478 

7.1154 

.8522 

.9903  .9958 

82°  00' 

1.4312 

.1425 

10 

.1421 

.1525 

.1435 

.1569 

6.9682 

.8431 

.9899  .9^)56 

50 

1.4283 

.1454 

20 

.1449 

.1612 

.1465 

.1658 

6.8269 

.8342 

.9894  .9954 

40 

1.4254 

.1484 

30 

.1478 

.1697 

.1495 

.1745 

6.6912 

.8255 

.9890  .9952 

30 

1.4224 

.1513 

40 

.1507 

.1781 

.1524 

.1831 

6.5(506 

.8169 

.9886  .9950 

20 

1.4195 

.1542 

50 

.1536 

.1863 

.1554 

.1915 

6.4348 

.8085 

.9881  .9948 

10 

1.4166 

.1571 

9°  00' 

.1564 

.1943 

.1584 

.1997 

6.3138 

.8003 

.9877  .9946 

81°  00' 

1.4137 

Value 

Logio 

Value 

Lo^io 

Value 

Logio 

Value   Logio 

Degrees 

Radians 

Cosine 

Cotangent 

Tangent 

Sine 

Four  Place  Trigonometric  Functions 


279 


[C 

liaracterist 

ics  of  Logarith 

ins  omitted  — 

determine  by  the  usual  rule  from  the  val 

ue] 

Radians 

Dkgrees 

Sine 

Tangent 

Cotangent 

Cosine 

V^alue 

Logio 

Value 

Lwgio 

Value 

L<>?io 

Value 

Lopio 

.1571 

9°  00' 

.1564 

.m3 

.1584 

.1997 

6..3138 

.8003 

.9877 

.9^)46 

81°  00' 

1.4137 

.1600 

10 

.1593 

.2022 

.1614 

.2078 

6.1970 

.7922 

.9872 

.9944 

50 

1.4108 

.1(>29 

20 

.1622 

.2100 

.1644 

.2158 

6.0844 

.7842 

.98(58 

.9942 

40 

1.4075» 

.1058 

30 

.1650 

.2176 

.1673 

.2236 

5.9758 

.7764 

.98(53 

.f)940 

30 

1.4050 

.1087 

40 

.1679 

.2251 

.1703 

.2313 

5.8708 

.7(587 

.9858 

.9938 

20 

1.4021 

.1716 

50 

.1708 

.2324 

.1733 

.2389 

5.7694 

.7611 

.9853 

.9936 

10 

1.3992 

.1745 

10°  00' 

.1736 

.2397 

.1763 

.2463 

5.6713 

.75.37 

.9848 

.9934 

80°  00' 

1.3963 

.1774 

10 

.1765 

.2468 

.1793 

.25.*5(5 

5.57(54 

.7464 

.9843 

.9931 

50 

1.3934 

.1804 

20 

.171^ 

.2538 

.1823 

.2609 

5.4815 

.7391 

.9838 

.9929 

40 

1.35)04 

.1833 

30 

.1822 

.2606 

.1853 

.2(580 

5.3955 

.7320 

.9833 

.9927 

30 

1.3875 

.1862 

40 

.1851 

.2674 

.1883 

.27.50 

5.3093 

.7250 

.9827 

.9924 

20 

1.3846 

.1891 

.50 

.1880 

.2740 

.1914 

.2819 

5.2257 

.7181 

.9822 

.9922 

10 

1.3817 

.1920 

11°00' 

.lf)08 

.280(5 

.1944 

.2887 

5.1446 

.7113 

.9816 

.9919 

79°  00' 

1.3788 

.1949 

10 

.1937 

.2870 

.1974 

.2953 

5.0(558 

.7047 

.9811 

.9917 

.50 

1.3759 

.1978 

20 

.UK)5 

.29;^ 

.2004 

.3020 

4.985)4 

.6980 

.9805 

.9<)14 

40 

1.3730 

.2007 

30 

.1994 

.2997 

.2035 

.3085 

4.9152 

.6915 

.9799 

.9912 

.30 

1..3701 

.2036 

40 

.2022 

.3058 

.2065 

.3149 

4.8430 

.6851 

.9793 

.9909 

20 

1.3672 

.20(>5 

50 

.2051 

.3119 

.2095 

.3212 

4.7729 

.6788 

.9787 

.9907 

10 

1.3643 

.2094 

12°  00' 

.2079 

.3179 

.2126 

.3275 

4.7046 

.6725 

.9781 

.9904 

78°  00' 

1.3614 

.2123 

10 

.2108 

.3238 

.2156 

.333(5 

4.6382 

.6664 

.9775 

.9901 

50 

1.3584 

.2153 

20 

.2136 

.329() 

.218() 

.3397 

4.5736 

.6(503 

.9769 

.9899 

40 

1.3555 

.2182 

30 

.21(>4 

.3353 

.2217 

.M.58 

4.5107 

.6542 

.9763 

.9896 

30 

1.3526 

.2211 

40 

.2193 

.3410 

.2247 

.3517 

4.4494 

.6483 

.9757 

.9893 

20 

1..3497 

.2240 

50 

.2221 

.34()6 

.2278 

.3576 

4.3897 

.6424 

.9750 

.9890 

10 

1.3468 

.2269 

13°  00 

.2250 

.3521 

.2309 

.3634 

4..3315 

.6366 

.9744 

.9887 

77° 00' 

1.3439 

.2298 

10 

.2278 

.3575 

.2339 

.3(591 

4.2747 

.6309 

.9737 

.9884 

50 

1.3410 

.2327 

20 

.2'M) 

.3629 

.2370 

.3748 

4.2193 

.(5252 

.9730 

.9881 

40 

1.3381 

.23.56 

30 

.2334 

.3()82 

.2401 

.3804 

4.1(553 

.619(5 

.9724 

.9878 

30 

1.33.52 

.2385 

40 

,2.363 

.37.">4 

.2432 

.3859 

4.1126 

.6141 

.9717 

.9875 

20 

1.3323 

.2414 

50 

.2391 

.3786 

.2462 

.3914 

4.0(511 

.(3086 

.9710 

.9872 

10 

1.325)4 

.2443 

14°  00' 

.2419 

.3837 

.2493 

.3^)08 

4.0108 

.6032 

.9703 

.9869 

76°  00' 

1..3265 

.2473 

10 

.2447 

..'{887 

.2524 

.4021 

3.9(517 

.5979 

.9696 

.9866 

50 

1.3235 

.2502 

20 

.2476 

.3937 

.2555 

.4074 

3.913(5 

.,5926 

.%89 

.9863 

40 

1.3206 

.2531 

30 

.2504 

.3986 

.2586 

.4127 

3.8(567 

.5S73 

.9(581 

.9859 

30 

1.3177 

.25(K) 

40 

.2.532 

.40:x5 

.2617 

.4178 

3.8208 

.5822 

.9674 

.985(5 

20 

1.3148 

.2589 

50 

.2560 

.4083 

.2648 

.4230 

3.7760 

.5770 

.9(567 

.9853 

10 

1.3119 

.2618 

15°00' 

.2588 

.41.30 

.2(579 

.4281 

3.7321 

.5719 

.9659 

.9&49 

76°00' 

1.3090 

.2647 

10 

.2()16 

.4177 

.2711 

.4331 

3.6891 

.5(5(5!) 

.9(552 

.9846 

50 

1.3061 

.2676 

20 

.2(i41 

.4223 

.2742 

.4381 

3.(5470 

.5619 

.9644 

.9843 

40 

1.3032 

.2705 

30 

.2672 

.4269 

.2773 

.44:30 

3  6059 

.5570 

.9():36 

.9839 

30 

1.3003 

.2734 

40 

.2700 

.4314 

.2805 

.4479 

3.565(5 

.5521 

.9628 

.9836 

20 

1.2974 

.2763 

50 

.2728 

,4359 

.2836 

.4527 

3.52(51 

.5473 

.9621 

.9832 

10 

1.2945 

.2793 

16°  00' 

.2756 

.4403 

.2867 

.4575 

3.4874 

.5425 

.9(513 

.9828 

74° 00' 

1.2915 

.2822 

10 

.2784 

.4447 

.289*) 

.4(522 

3.4495 

..5378 

.9605 

.9825 

50 

1.2886 

.2851 

20 

.2812 

.4491 

.2931 

.4(5(59 

3.4124 

..5331 

.959(5 

.9821 

40 

1.28.57 

.2880 

30 

.2.S40 

.4533 

.2962 

.4716 

3.3759 

.5284 

.9588 

.9817 

30 

1.2828 

.2909 

40 

.28(38 

.457(> 

.25)94 

.4762 

3.3402 

.52.38 

.9580 

.9814 

20 

1.2799 

.2938 

50 

.28i)6 

.4618 

.3026 

.4808 

3.3052 

.5192 

.9572 

.9810 

10 

1.2770 

.2967 

17°  00' 

.2924 

.4659 

.;3057 

.48.-13 

3.2709 

.5147 

.9563 

.9806 

73°  00' 

1.2741 

.2996 

10 

.2952 

.4700 

.3(J89 

.4898 

3.2371 

.5102 

.9555 

.9802 

50 

1.2712 

.3025 

20 

.2979 

.4741 

..3121 

.4943 

3.2041 

.5057 

.9546 

.9798 

40 

1.2683 

.3054 

30 

.3007 

4781 

.3153 

.4987 

3.1716 

.5013 

.9537 

.9794 

30 

1.2(354 

..•3083 

40 

.3035 

.4821 

.3185 

.5031 

3.1397 

.4i)(59 

.9528 

.975)0 

20 

1.2(325 

.3113 

50 

.3062 

.4861 

.3217 

.5075 

3.1084 

.4925 

.9520 

.978(5 

10 

1.2595 

.3142 

18° 00' 

..3090 

.4900 

.3249 

.5118 

3.0777 

.4882 

.9511 

.9782 

72°  00' 

1.2566 

Value 

Logio 

Value 

LocTio 

Value 

Logio 

Value 

Logio 

Degrees 

Radians 

Cos 

NB 

Cotangent 

'I'ang 

ENT 

Sine 

280 


Four  Place  Trigonometric  Functions 


[Ctiaractoristics  of  Logarith 

ms  omitted  — 

ietermine  by  the  usual  rule  from  the  value] 

Radians 

Degkees 

Sine 

Tangent 

Cotangent 

Cosine 

Value 

Logio 

Value 

Logio 

Value 

Logio 

Value 

Logio 

.8142 

18°  00' 

.3090 

.4900 

.3249 

.5118 

3.0777 

.4882 

.9511 

.9782 

72°  00' 

1.2566 

.8171 

10 

.3118 

.4939 

.3281 

.5161 

3.0475 

.4839 

.9502 

.9778 

50 

1.2537 

.8JU0 

20 

.3145 

.4977 

.3314 

.5203 

3.0178 

.4797 

.9492 

.5)774 

40 

1.2508 

.3229 

30 

.3173 

.5015 

.3346 

.5245 

2.9887 

.4755 

.9483 

.9770 

30 

1.2479 

.3258 

40 

.3201 

.5052 

.3378 

.5287 

2.9600 

.4713 

.9i74 

.9765 

20 

1.2450 

.3287 

50 

.3228 

.5090 

.3411 

.5329 

2.9319 

.4671 

.9465 

.9761 

10 

1.2421 

.3316 

19°  00' 

.3256 

.5126 

.3443 

.5370 

2.9042 

.4630 

.9455 

.9757 

71°  00' 

1.2392 

.3.345 

10 

.3283 

.5163 

.3476 

.5411 

2.8770 

.4589 

.9446 

.9752 

50 

1.2363 

.3374 

20 

.3311 

.5199 

.3508 

.5451 

2.8502 

.4549 

.9436 

.9748 

40 

1.2334 

.3403 

30 

.3338 

.52.35 

.3541 

.5491 

2.8239 

.4509 

.9426 

.9743 

30 

1.2305 

.3432 

40 

.3365 

.5270 

.3574 

.5531 

2.7980 

.4469 

.9417 

.9739 

20 

1.2275 

.3462 

50 

.3393 

.5306 

.3607 

.5571 

2.7725 

.4429 

.9407 

.9734 

10 

1.2246 

.3491 

20°  00' 

.3420 

.5341 

..3640 

.5611 

2.7475 

.4389 

.9397 

.9730 

70°  00' 

1.2217 

.3520 

10 

.3448 

.5375 

.3673 

.5650 

2.7228 

.4350 

.9387 

.9725 

50 

1.2188 

.3.549 

20 

.;^75 

.5409 

.370(5 

.5689 

2.6985 

.4311 

.9377 

.9721 

40 

1.2159 

.3578 

30 

.3502 

.5443 

.3739 

.5727 

2.6746 

.4273 

.9367 

.9716 

30 

1.2130 

.3607 

40 

.3529 

.5477 

.3772 

.5766 

2.6511 

.42M 

.9356 

.9711 

20 

1.2101 

.3636 

50 

.3557 

.5510 

.3805 

.5804 

2.6279 

.4196 

.9346 

.9706 

10 

1.2072 

.3665 

21°  00' 

.3584 

.5543 

.3839 

.5842 

2.6051 

.4158 

.9336 

.9702 

69° 00' 

1.2043 

.3694 

10 

.3611 

.5576 

.3872 

.5879 

2.5826 

.4121 

.9325 

.9697 

50 

1.2014 

.3723 

20 

.3fi38 

.5609 

.3906 

.5917 

2.5605 

.4083 

.9315 

.9692 

40 

1.1985 

.3752 

30 

.3(565 

.5(541 

.3939 

.5954 

2.5386 

.4046 

.9304 

.9687 

30 

1.1956 

.3782 

40 

.3692 

.5()73 

.3973 

.5991 

2.5172 

.4009 

.9293 

.9682 

20 

1.1926 

.3811 

50 

.3719 

.57(J4 

.4006 

.6028 

2.4960 

.3972 

.9283 

.9(577 

10 

1.1897 

.3840 

22°  00' 

.3746 

.5736 

.4040 

.6064 

2.4751 

.3936 

.9272 

.%72 

68°  00' 

1.1868 

.3869 

10 

.3773 

.57(57 

.4074 

.6100 

2.4545 

.3900 

.9261 

.i)667 

50 

1.1839 

.3898 

20 

.3800 

.5798 

.4108 

.6136 

2.4342 

.3864 

.9250 

smi 

40 

1.1810 

.3927 

30 

.3827 

.5828 

.4142 

.6172 

2.4142 

.3828 

.9239 

.9(556 

30 

1.1781 

.3956 

40 

.3854 

.5859 

.4176 

.6208 

2.3945 

.3792 

.9228 

.9651 

20 

1.1752 

.3985 

50 

.3881 

.5889 

.4210 

.6243 

2.3750 

.3757 

.9216 

.9646 

10 

1.1723 

.4014 

23°  00' 

.3907 

.5919 

.4245 

.6279 

2.3559 

.3721 

.9205 

.9640 

67°  00' 

1.1694 

.4043 

10 

.39,-^ 

.5948 

.4279 

.6314 

2.3369 

.3686 

.9194 

.9635 

50 

1.1(565 

.4072 

20 

.35)61 

.5978 

.4314 

.6348 

2.3183 

.3652 

.9182 

.9629 

40 

1.1(536 

.4102 

30 

.3987 

.(5007 

.4;M8 

.6383 

2.2998 

.3617 

.9171 

.9624 

30 

1.160(5 

.4131 

40 

.4014 

.6036 

.4:383 

.(5417 

2.2817 

.3583 

.9159 

.9618 

20 

1.1577 

.4160 

50 

.4041 

.6065 

.4417 

.6452 

2.2637 

.3548 

.9147 

.9613 

10 

1.1548 

.4189 

24° 00' 

.4067 

.6093 

.4452 

.6486 

2.2460 

.3514 

.9135 

.9607 

66°  00' 

1.1519 

.4218 

10 

.4094 

.6121 

.4487 

.6520 

2  2286 

.3480 

.9124 

.9602 

50 

1.1490 

.4247 

20 

.4120 

.6149 

.4522 

.6553 

2.2113 

.3447 

.9112 

.9596 

40 

1.14(51 

.4276 

30 

.4147 

.6177 

.4557 

.6587 

2.1943 

.3413 

.9100 

.9590 

30 

1.1432 

.4305 

40 

.4173 

.6205 

.4592 

.6620 

2.1775 

.3380 

.9088 

.9584 

20 

1.1403 

.4334 

50 

.4200 

.6232 

.4628 

.6654 

2.1609 

.3346 

.9075 

.9579 

10 

1.1374 

.4363 

25°  00' 

.4226 

.6259 

.4663 

.6687 

2.1445 

.3313 

.9063 

.9573 

65°  00' 

1.1345 

.4392 

10 

.4253 

.6286 

.4699 

.6720 

2.1283 

.3280 

.9051 

.9567 

50 

1.1316 

.4422 

20 

.4279 

.6313 

.4734 

.6752 

2.1123 

.3248 

.9038 

.9561 

40 

1.1286 

.4451 

30 

.4305 

.6;mo 

.4770 

.6785 

2.0965 

.3215 

.9026 

.9555 

30 

1.1257 

.4480 

40 

.4331 

.636(5 

.4806 

.6817 

2.0809 

.3183 

.9013 

.9549 

20 

1.1228 

.4509 

50 

.4358 

.6392 

.4841 

.6^50 

2.0655 

.3150 

.9001 

.9543 

10 

1.1195) 

.4538 

26°  00' 

.4384 

.6418 

.4877 

.6882 

2.0503 

.3118 

.8988 

.9537 

64°  00' 

1.1170 

.4567 

10 

.4410 

.6444 

.4913 

.6914 

2.0353 

.3086 

.8975 

.9530 

50 

1.1141 

.4596 

20 

.4436 

.6470 

.4950 

.(5946 

2.0204 

.3054 

.8962 

.9524 

40 

1.1112 

.4625 

30 

.4462 

.6495 

.4986 

.(5977 

2.0057 

.3023 

.8949 

.9518 

30 

1.1083 

.4654 

40 

.4488 

.6521 

.5022 

.7009 

1.9912 

.2991 

.893(5 

.9512 

20 

1.1054 

.4(583 

50 

.4514 

.6546 

.5059 

.7040 

1.9768 

.2960 

.8923 

.9505 

10 

1.1025 

.4712 

27°  00' 

.4540 

.6570 

.5095 

.7072 

1.9626 

.2928 

.8910 

.&499 

63°  00' 

1.0996 

Value 

Logio 

Value 

Lf>Pio 

Value 

Logio 

Value 

Logio 

Degrees 

Radians 

Cosine 

Cotangent 

Tangent 

Sine 

Four  Place  Trigonometric  Functions 


281 


[Characteristics  of  Logarithms  omi 

tted  —  determine  by  the  usual 

rule  from  the  val 

le] 

Radians 

Deobees 

Sine 

Tangent 

Cotangent 

Cosine 

Value 

Logio 

Value 

Logio 

Value 

Logio 

Value 

Logio 

.4712 

27°  00' 

.4540 

.6570 

.5095 

.7072 

1.9(326 

.2928 

.8910 

.9499 

63°  00' 

1.0996 

.4741 

10 

.45(36 

.6595 

.5132 

.7103 

1.9486 

.2897 

.8897 

.9492 

50 

1.09(56 

.4771 

20 

.4592 

.6()20 

.51(59 

.7134 

1.9347 

.28(36 

.8884 

.948(5 

40 

1.0937 

.4800 

30 

.4617 

.(3(344 

.5206 

.7165 

1.9210 

.2835 

.8870 

.9479 

30 

1.0908 

.4829 

40 

.4643 

.m>s 

.5243 

.7196 

1.9074 

.2804 

.8857 

.9473 

20 

1.0879 

.4858 

50 

.4669 

.6692 

.5280 

.7226 

1.8940 

.2774 

.8843 

.9466 

10 

1.0850 

.4887 

28°  00' 

.4695 

.6716 

.5317 

.72.57 

1.8807 

.2743 

.8829 

.9459 

62°  00' 

1.0821 

.4916 

10 

.4720 

.()740 

.5354 

.7287 

1.8(576 

.2713 

.8816 

.9453 

50 

1.0792 

.4945 

20 

.4746 

.(3763 

.5392 

.7317 

1.8546 

.2683 

.8802 

.944(5 

40 

1.07(53 

.4974 

30 

.4772 

.0787 

..5430 

.7348 

1.8418 

.2652 

.8788 

.9439 

.'30 

1.07.34 

.5003 

40 

.4797 

.(3810 

.54(37 

.7378 

1.8291 

.2(522 

.8774 

.9432 

20 

1.0705 

.5032 

50 

.4823 

.(5833 

.5505 

.7408 

1.8165 

.2592 

.8760 

.9425 

10 

1.0676 

.5061 

29°  00' 

.4848 

.(3856 

.5543 

.74:38 

1.8040 

.2.5(52 

.8746 

.9418 

61°  00' 

1.0647 

.5091 

10 

.4874 

.6878 

..5581 

.74(57 

1.7917 

.253:3 

.87.32 

.9411 

50 

1.0(517 

.5120 

20 

.48«)«) 

.(5901 

.5(319 

.7497 

1.775)6 

.2503 

.8718 

.9404 

40 

1.0588 

.5149 

30 

.4924 

.()923 

..5(558 

.752(3 

1.7(575 

.2474 

.8704 

.9397 

:30 

1.0.559 

.5178 

40 

.4f)50 

.&m 

.5(19(3 

.75rAi 

1.7.5.56 

.2444 

.8(589 

.9390 

20 

1.05:30 

.5207 

50 

.4975 

.69(38 

.5735 

.7585 

1.7437 

.2415 

.8675 

.9383 

10 

1.0501 

.623(3 

30°  00' 

liOOO 

.(5990 

.5774 

.7614 

1.7:321 

.2.38(5 

.8(560 

.9.375 

60°  00' 

1.0472 

.52(>5 

10 

.5025 

.7012 

.5812 

.7(544 

1.7205 

.2:3,5(5 

.8(546 

.9:368 

50 

1.0443 

.5294 

20 

.5050 

.70.'3;3 

.5851 

.7(573 

1.7090 

.2:527 

.8(331 

.9:3(31 

40 

1.0414 

.5323 

30 

.5075 

.70.55 

.5890 

.7701 

1.(5977 

.2299 

.8(316 

.9353 

30 

1.0385 

.5352 

40 

.5100 

.707(5 

.59;J0 

.77:30 

1.(58(54 

.2270 

.8(301 

.9.34(5 

20 

1.0356 

.5381 

50 

.5125 

.7097 

.5969 

.7759 

1.6753 

.2241 

.8587 

.9338 

10 

1.0327 

.Mil 

31°  00' 

MrrO 

.7118 

.600<) 

.7788 

1.(5643 

.2212 

.8572 

.9331 

59° 00' 

1.0297 

.5440 

10 

.5175 

.7139 

.(3048 

.781(5 

1  .(55:34 

.2184 

.8557 

.9323 

50 

1.02(58 

.54()9 

20 

.5200 

.71(50 

.(5088 

.7845 

1.(5426 

.2155 

.8.542 

.9315 

40 

1.02.39 

.5498 

30 

.5225 

.7181 

.(5128 

.787:5 

1.6319 

.2127 

.8.526 

.9:308 

30 

1.0210 

.5527 

40 

.52.50 

.7201 

.61(58 

.75)02 

1.(3212 

.2098 

.8511 

.9:300 

20 

1.0181 

.5556 

50 

.5275 

.7222 

.6208 

.7930 

1.6107 

.2070 

.8496 

.9292 

10 

1.0152 

.5585 

32°  00' 

.5299 

.7242 

.6249 

.7958 

1.6003 

.2042 

.8480 

.9284 

58°  00' 

1.0123 

.5<)14 

10 

.5324 

.72(52 

.(5289 

.798(5 

1.5900 

.2014 

.84(35 

.927(3 

50 

1.00514 

.5643 

20 

.5:548 

.7282 

.6.330 

.8014 

1.5798 

.198(5 

.8450 

.9268 

40 

1.00(55 

.5(i72 

30 

.5373 

.7302 

.6371 

.8042 

1.5(397 

.1958 

.8434 

.92(30 

30 

1.0036 

.5701 

40 

.5398 

.7322 

m\2 

.8070 

1.5,597 

.19:30 

.8418 

.9252 

20 

1.0007 

.5730 

50 

.5422 

.7342 

.6453 

.8097 

1.5497 

.1903 

.8403 

.9244 

10 

.9977 

.5760 

33°  00' 

.5446 

.73(51 

.6494 

.8125 

1.5399 

.1875 

.8387 

.92.36 

57°  00' 

.9948 

.5789 

10 

.5471 

.7380 

.(3536 

.8153 

1.5301 

.1847 

.8371 

.9228 

50 

.95119 

.5818 

20 

.5495 

.7400 

.(5577 

.8180 

1.5204 

.1820 

.8355 

.9219 

40 

.9890 

.5847 

30 

.5519 

.7419 

.(5(519 

.8208 

1.5108 

.1792 

.8339 

.9211 

30 

.9861 

.5876 

40 

.5544 

.7438 

.(5(5(51 

.82:35 

1.5013 

.1765 

.8323 

.920:3 

20 

.9832 

.5905 

50 

.5568 

.7457 

.(5703 

.8263 

1.4919 

.1737 

.8307 

.9194 

10 

.9803 

.5934 

34°  00' 

.5592 

.7476 

.6745 

.82^)0 

1.4826 

.1710 

.8290 

.9186 

56°  00' 

.9774 

.5963 

10 

.5616 

.7494 

.(5787 

.8317 

1.47:33 

.1683 

.8274 

.9177 

50 

.9745 

.5992 

20 

.5(m 

.7513 

.68:50 

.8344 

1.4641 

.1(556 

.8258 

.91(59 

40 

.9716 

.6021 

30 

.5(3(34 

.7.531 

.6873 

.8:371 

1.4550 

.1(329 

.8241 

.91(30 

30 

.9687 

.6050 

40 

.5(388 

.75;-30 

.6916 

.8398 

1.44(50 

.1602 

.8225 

.9151 

20 

.9(357 

.(3080 

50 

.5712 

.7568 

.6959 

.8425 

1.4370 

.1575 

.8208 

.9142 

10 

.9628 

.6109 

35° 00' 

.5736 

.7586 

.7002 

.8452 

1.4281 

.1548 

.8192 

.9134 

55°  00' 

.9599 

.6138 

10 

.57(50 

.7(304 

.7046 

.8479 

1.4193 

.1521 

.8175 

.9125 

.50 

.9570 

.6167 

20 

..5783 

.7(322 

.7089 

.8506 

1.4106 

.1494 

.8158 

.9116 

40 

.9541 

.6196 

30 

.5807 

.7(540 

.71:33 

.8533 

1.4019 

.1467 

.8141 

.9107 

30 

.9512 

.6225 

40 

.5831 

.7(357 

.7177 

.8559 

1.39:34 

.1441 

.8124 

.9098 

20 

.9483 

.6254 

50 

.5854 

.7675 

.7221 

.8586 

1.3848 

.1414 

.8107 

.9089 

10 

.9454 

.6283 

36°  00' 

.5878 

.7692 

.7265 

.8613 

1.3764 

.1387 

.8090 

.9080 

54°  00' 

.9425 

Value 

Log,o 

Value 

Lopio 

Value 

Logio 

Value 

Logio 

Degbees 

Uadians 

Cosine 

Cotangent 

Tangent 

Sine 

282 


Four  Place  Trigonometric  Functions 


[Characteristics  of  Logarithms  omitted  —  determine  by  the  usual  rule  from  the  val 

uel 

Radians 

Degbeee 

Sine 

Tangent 

Cotangent 

Cosine 

Value  Logio 

Value   Logio 

Value   Logio 

Value  Logio 

.6283 

36°  00' 

.5878  .7692 

.7265  .8613 

1.3764  .1387 

.8090  .9080 

54°  00' 

.9425 

.6312 

10 

.5901  .7710 

.7310  .8639 

1.3680  .13(51 

.8073  .9070 

50 

.9396 

.6341 

20 

.5925  .7727 

.7355  .86(56 

1.3597  .13(54 

.8056  .90(51 

40 

.9367 

.6370 

30 

.5948  .7744 

.7400  .8692 

1.3514  .1308 

.8039  .9052 

30 

.9338 

.6400 

40 

.5972  .7761 

.7445  .8718 

1.34.32  .1282 

.8021  .9042 

20 

.9308 

.6429 

50 

.5995  .7778 

.7490  .8745 

1.3351  .1255 

.8004  .9033 

10 

.9279 

.6458 

37°  00' 

.6018  .7795 

.7536  .8771 

1.3270  .1229 

.7986  .9023 

63°  00' 

.9250 

.6487 

10 

.6041  .7811 

.7581  .8797 

1.3190  .1203 

.7i)(59  .9014 

50 

.9221 

.6516 

20 

.(5065  .7828 

.7627  .8824 

1.3111  .1176 

.7951  .9004 

40 

.9192 

.6545 

30 

.6088  .7844 

.7(573  .8850 

1.3032  .1150 

.7934  .89<)5 

30 

.9163 

.6574 

40 

.6111  .7861 

.7720  .8876 

1.2954  .1124 

.7916  .8985 

20 

.9134 

.6(503 

50 

.6134  .7877 

.7766  .8902 

1.2876  .1098 

.7898  .8975 

10 

.9105 

.6632 

38°  00' 

.6157  .7893 

.7813  .8928 

1.2799  .1072 

.7880  .8965 

52°  00' 

.9076 

.66(51 

10 

.6180  .7910 

.78(50  .8954 

1.2723  .1046 

.7862  .8955 

50 

.IKMT 

.6690 

20 

.6202  .7926 

.7907  .8980 

1.2647  .1020 

.7844  .8945 

40 

.9018 

.6720 

30 

.6225  .7941 

.79.54  .9006 

1.2572  .0994 

.7826  .8935 

30 

.8988 

.6749 

40 

.6248  .7957 

.8002  .90,32 

1.2497  .0968 

.7808  .8925 

20 

.8959 

.6778 

50 

.6271  .7973 

.8050  .9058 

1.2423  .0942 

.7790  .8915 

10 

.8930 

.6807 

39°  00' 

.6293  .7989 

.8098  .9084 

1.2349  .0916 

.7771  .8905 

51°00' 

.8901 

.68.36 

10 

.631(5  .8004 

.8146  .9110 

1.2276  .0890 

.77.53  .8895 

50 

.8872 

.68(55 

20 

.6338  .8020 

.8195  .9i;55 

1.2203  .08tJ5 

.7735  .8884 

40 

.8843 

.6894 

30 

.(5.3(51  .80;55 

.8243  .91()1 

1.2131  .0839 

.7716  .8874 

30 

.8814 

.6923 

4(J 

.6383  .80.50 

.8292  .9187 

1.20.59  .0813 

.7698  .8864 

20 

.8785 

.6952 

50 

.6406  .8066 

.8342  .9212 

1.1983  .0788 

.7679  .8853 

10 

.8756 

.6981 

40°  00' 

.6428  .8081 

.8391  .9238 

1.1918  .0762 

.7660  .8843 

50°  00' 

.8727 

.7010 

10 

.6450  .som 

.8441  .9264 

1.1847  .07:5(5 

.7(542  .8832 

50 

.8698 

.7039 

20 

.6472  .8111 

.8491  .9289 

1.1778  .0711 

.7623  .8821 

40 

.8668 

.7069 

30 

.(M94  .8125 

.8541  .9315 

1.1708  .0685 

.7604  .8810 

30 

.8639 

.7098 

40 

.6517  .8140 

.8591  .9:341 

1.1640  .0659 

.7585  .8800 

20 

.8610 

.7127 

50 

.6539  .8155 

.8642  .9366 

1.1571  .0634 

.7566  .8789 

10 

.8581 

.7156 

41°00' 

.6561  .8169 

.8(593  .9392 

1.1504  .0608 

.7547  .8778 

49°  00' 

.8552 

.7185 

10 

.6583  .8184 

.8744  .<U17 

1.1436  .0583 

.7528  .8767 

50 

.&523 

.7214 

20 

.6()04  .8198 

.8796  .9443 

1.13(59  .0557 

.7509  .8756 

40 

.8494 

.7243 

30 

.6626  .8213 

.8847  .9468 

1.1303  .0532 

.74<>0  .8745 

30 

.8465 

.7272 

40 

.6648  .8227 

.88^)9  .94M 

1.1237  .0506 

.7470  .8733 

20 

.8436 

.7301 

50 

.6670  .8241 

.8952  .9519 

1.1171  .0481 

.7451  .8722 

10 

.8407 

.7330 

42°  00' 

.66<)1  .8255 

.9004  .9544 

1.1106  .0456 

.7431  .8711 

48°  00' 

.8378 

.7359 

10 

.6713  .8269 

.9057  .9570 

1.1041  .0430 

.7412  .86{)9 

50 

.8348 

.7389 

20 

.6734  .8283 

.9110  .9595 

1.0977  .0405 

.7392  .8688 

40 

.8319 

.7418 

30 

.6756  .8297 

.9163  .9(521 

1.0913  .0379 

.7373  .8676 

30 

.8290 

.7447 

40 

.6777  .8311 

.9217  .9646 

1.0850  .0354 

.7353  .86(55 

20 

.8261 

.7476 

50 

.(5799  .8324 

.9271  .9671 

1.0786  .0329 

.7333  .8653 

10 

.8232 

.7505 

43°  00' 

.6820  .8338 

.9325  .9697 

1.0724  .0303 

.7314  .8641 

47°  00' 

.8203 

.7534 

10 

.(5841  .8351 

.9380  .9722 

1.0661  .0278 

.7294  .8629 

50 

.8174 

.7563 

20 

.6862  .8365 

.9435  .9747 

1.059<)  .0253 

.7274  .8(518 

40 

.8145 

.7592 

30 

.6884  .8378 

.9490  .9772 

1.0538  .0228 

.7254  .8606 

30 

.8116 

.7621 

40 

.6^)05  .8391 

.9545  .9798 

1.0477  .0202 

.7234  .8594 

20 

.8087 

.7650 

50 

.6926  .8405 

.9601  .9823 

1.0416  .0177 

.7214  .8582 

10 

.8058 

.7679 

44°  00' 

.6947  .8418 

.9657  .9848 

1.0.355  .0152 

.7193  .8569 

46°  00' 

.8029 

.7709 

10 

.6%7  .8431 

.9713  .9874 

1.0295  .0126 

.7173  .8557 

50 

.79i)9 

.7738 

20 

.6988  .8444 

.9770  .98^)9 

1.0235  .0101 

.7153  .8545 

40 

.7970 

.7767 

30 

.7009  .8457 

.9827  .9924 

1.0176  .0076 

.7133  .85:32 

30 

.7941 

.7796 

40 

.7030  .84(i9 

.9884  .9949 

1.0117  .0051 

.7112  .8520 

20 

.7912 

.7825 

50 

.7050  .8482 

.9942  .9975 

1.0058  .0025 

.7092  .8507 

10 

.7883 

.7854 

45°  00' 

.7071  .8495 

1.0000  .0000 

1.0000  .0000 

.7071  .8495 

45°  00' 

.7854 

Value  Logio 

Value   Logio 

Value   Logio 

Value  Logio 

Degrees 

Radians 

Cosine 

Cotangent 

Tangent 

Sine 

Table  X  — Compound  Interest:  (l+r)»  283 

Amount  of  One  Dollar  Principal  at  Compound  Interest  After  n  Years 


n 

2<fo 

2h^o 

3^0 

31% 

4% 

4^% 

5% 

O'/o 

7% 

1 
2 
3 

4 

5 
6 

7 
8 
9 

1.0200 
1.0404 
1.0612 

1.0824 
1.1041 
1.1262 

1.1487 
1.1717 
1.19.51 

1.0250 
1.050() 
1.0769 

1.1038 
1.1314 
1.1597 

1.1887 
1.2184 
1.2489 

1.0300 
1.0609 
1.0927 

1.1255 
1.1593 
1.1941 

1.2299 
1.2668 
1.3048 

1  0350 
1.0712 
1.1087 

1.1475 
1.1877 
1.2293 

1.2723 
1.3168 
l..%29 

1.0400 
1.0816 
1.1249 

1.1699 
1.2167 
1.2653 

1.3159 
1.3686 
1.4233 

1.0450 
1.0920 
1.1412 

1.1925 
1.2462 
1.3023 

1.3609 
1.4221 
1.4861 

1.0500 
1.1025 
1.1576 

1.2155 
1.2763 
1.3401 

1.4071 
1.4775 
1.5513 

1.0600 
1.12.36 
1.1910 

1.2625 
1.3382 
1.4185 

1.50156 
1.5938 
1.6895 

1.0700 
1.1449 
1.2250 

1.3108 
1 .4026 
1.5007 

1 .6058 
1.7182 

1 .8385 

l.%72 

2.1049 
2.2522 
2.4098 

2.5785 
2.7590 
2.9522 

3.1588 
3.379t) 
3(51(55 

10 

1.219C 

1.2801 

1.34.39 

1.4106 

1.4802 

1.5.530 

1.6289 

1.7908 

11 
12 
13 

14 
15 
16 

17 

18 
19 

1.2434 
1.2682 
1.2936 

1.3195 
l.;U59 
1.3728 

1.4002 
1.4282 
1.45()8 

1.3121 
1.3449 
1.3785 

1.4130 
1.4483 
1.4845 

1.5216 

1.5597 
1.5987 

1.3842 
1.4258 
1.4685 

1.5126 
1.5580 
1.6047 

1.6528 
1.7024 
1.7535 

1.4<300 
1.5111 
1.5040 

1.6187 
l.<)753 
1.7340 

1.7947 
1.8575 
1.9225 

1.5.395 
1.(^10 
1.6651 

1.7317 

1.8009 
1.8730 

1.9479 
2.0258 
2.1068 

1.6229 
1.69.59 
1.7722 

1.8519 
1.93.53 
2.0224 

2.1134 

2.2085 
2.3079 

1.7103 
1.7959 
1.8856 

1.9799 
2.0789 
2.1829 

2.2920 
2.40(M) 
2..5270 

1.8983 
2.0122 
2.1329 

2.2609 
2..39(K5 
2.5404 

2.6928 
2.8543 
3.025(5 

20_ 

21 

22 
23 

24 

25 
26 

27 
28 
29 

1.48.59 

1.6386 

1.80(>1 

1.9898 

2.1911 

2.4117 

2.6533 

3.2071 

3.8697 

1.5157 
1.54(K) 
1.5769 

1.6084 
l.(;40<) 
1.6734 

1.7069 
1.7410 
1.7758 

1.6796 
1.721() 
1.7646 

1.8087 
1.85.39 
1.9003 

1.9478 
1 .9«)65 
2.04('4 

1.8603 
1.9161 
1.9736 

2.0328 

2.mm 

2.1566 

2.2213 

2.2879 
2..35<;() 

2.0594 
2.1.315 
2.2061 

2.2833 
2.;3()32 
2.44(X) 

2.5316 
2.6202 
2.7119 

2.2788 
2.3699 
2.4(J47 

2.5633 
2.6()58 
2.7725 

2.88.34 
2.9987 
3.1187 

2.5202 
2.6.337 
2.7522 

2.8760 
3.0054 
3.1407 

3.2820 
3.4297 
3.5840 

2.7860 
2.92.53 
3.0715 

3.2251 
3.3864 
3.5557 

3.7335 
3.9201 
4.1161 

3.3996 
3.6035 
3.8197 

4.0489 
4.2919 
4.5494 

4.8223 
5.1117 

5.4184 

4.1406 
4.4304 
4.7405 

5.0724 
5.4274 
5.8074 

6.21.39 

6.(5488 
7.1143 

30 

1.8114 

2.0976 

2.4273 

2.8068 

3.24:u 

3.7453 

4.3219 

5.74^5 

7.6123 

31 
32 
33 

34 

35 
36 

37 
38 
39 

1 .8476 
1.S845 
1.9222 

1.9607 
1.9999 
2.0399 

2.0807 
2.1223 
2.1647 

2.1500 
2.2038 
2.2589 

2.3153 
2.3732 
2.4325 

2.4933 
2.5557 
2.6196 

2.5001 
2.5751 
2.6523 

2.7319 

2.81.39 
2.8983 

2.9852 
3.0748 
3.1670 

2.90.50 
3.0067 
3.1119 

3.2209 
3.3336 
3.4503 

3.5710 
3.6960 
3.8254 

3.3731 

3.5081 
3.6484 

3.7943 
3.9461 
4.1039 

4.2681 
4.4388 
4.6164 

3.9139 
4.0900 
4.2740 

4.46(4 
4.6673 

4.8774 

5.0969 
5.3262 
5.5659 

5.8164 

4.5380 
4.7649 
5.0032 

5.2533 
5.51(50 
5.7918 

6.0814 
6.3855 
6.7048 

6.0881 
6.4.5.34 
6.8406 

7.2510 
7.(5861 
8.1473 

8.6361 
9.1543 

9.7035 

8.1451 
8.7153 
9.3253 

9.9781 
10.6766 
11.4239 

12.22,36 
13!0793 
13.9948 

40 

2.2080 

2.6851 

3.2620 

3.9.593 

4.8010 

7.0400 

10.28.57 

10.9029 
11.5.570 
12.2505 

12.9855 
13.7(546 
14.5905 

15.4659 
16.3939 
17.3775 

14.9745 

41 
42 
43 

44 
45 

46 

47 
48 
49 

60 

2.2522 
2.2972 
2.3432 

2.3901 
2.4379 

2.4866 

2.5.363 

2.5871 
2.6388 

2.7522 
2.8210 
2.8915 

2.9638 
3.0379 
3.1139 

3.1917 
3.2715 
3.353:i 

3.3599 
3.4607 
3.5645 

3.6715 
3.7816 
3.8950 

4.0119 
4.1323 
4.25(>2 

4.0978 
4.2413 
4.3897 

4.5433 
4.7024 
4.8669 

5.0,373 
5.2136 
5.3961 

5.5849 

4.9931 
5.1928 
5.4005 

5.6165 
5.8412 
6.0748 

6..3178 
6.5705 
6.8333 

7.1067 

6.0781 
6.3516 
6.6374 

6.9361 
7.2482 
7.5744 

7.9153 

8.2715 
8.«i437 

9.0326 

7.3920 
7.7616 
8.1497 

8.5572 
8.98.50 
9.4343 

9.1XXK) 
10.4013 
10.9213 

11.4(574 

16.0227 
17.1443 
18.3444 

19.6285 
21.0025 
22.4726 

24.0457 

25.7280 
27.5299 

29.4576 

2.6916 

3.4371 

4.3839 

18.4202 

284  Table  XI  —  Compound  Discount:  1/(1  +  r)^ 

Present  Value  of  One  Dollar  Due  at  the  End  of  n  Years 


n 

2^/0 

^h'/o 

3% 

31^0 

4% 

4^</c 

5% 

6% 

7% 

1 

2 
3 

4 
5 
6 

7 
8 
9 

10 

11 

12 
13 

14 
15 
16 

17 

18 
19 

20 

.98039 
.96117 
.94232 

.92385 
.90573 

.88797 

.87056 
.85349 
.83(576 

.97561 
.95181 
.928(50 

.90595 
.88385 
.86230 

.84127 
.82075 
.80073 

.97087 
.94260 
.91514 

.88849 
.86261 
.83748 

.81309 
.78941 
.76642 

.96618 
.93351 
.90194 

.87144 
.84197 

.81350 

.78599 
.75941 
.73373 

.96154 
.92456 
.88900 

.85480 
.82103 
.79031 

.75992 
.73069 
.70259 

.95694 
.91573 
.87630 

.83856 
.80245 
.76790 

.73483 
.70319 
.67200 

.95238 
.90703 
.86384 

.82270 
.78353 
.74622 

.71068 
.67684 
.64461 

.94340 
.89000 
.83962 

.79209 
.74726 
.70496 

.66506 
.(32741 
.59190 

.93458 
.87341 
.81630 

.7(5290 
.71299 
.66634 

.62275 
.58201 
.54393 

.82035 

.78120 

.74409 

.70802 

.67556 

.64393 

.61391 

.55839 

.50835 

.8042(5 
.78849 
.77303 

.75788 
.74301 
.72845 

.71416 
.70016 
.(W643 

.76214 
.74.356 
.72542 

.70773 
.(59017 
.673(52 

.65720 
.64117 
.62553 

.72242 

.70138 
.68095 

.66112 
.64186 
.(52317 

.60502 
.58739 
.57029 

.68405 
.(56178 
.63940 

.61778 
.59689 
.57671 

.55720 
.53836 
.52016 

.64958 
.62460 
.60057 

.57748 
.55526 
.53391 

.51337 
.493(53 
.47461 

.61620 
.5896(5 
.56427 

.53997 
.51(572 
.49447 

.47318 
.45280 
.43330 

.58468 
.55684 
.53032 

.50507 
.48102 
.45811 

.43630 
.41552 
.39573 

.52679 
.49607 
.46884 

.44230 
.41727 
.39365 

.37136 
.350.-54 
.33051 

.47509 
.44401 
.41496 

.38782 
.36245 
.33873 

.31657 
.29586 
.27651 

.67297 

.61027 

.553(58 

.50257 

.456.39 

.41464 

.37689 

.31180 

.29416 
.27751 
.26180 

.24698 
.23300 
.21981 

.20737 
.19563 
.18456 

.25842 

21 
22 
23 

24 
25 
26 

27 

28 
29 

.65978 
.64684 
.63416 

.62172 
.60953 
.59758 

.58586 
.57437 
.56311 

.59539 
.58086 
.56670 

.55288 
.53939 
.52623 

.51340 

.50088 
.488(5(5 

.53755 
.52189 
.50669 

.49193 
.47761 
.46369 

.45019 
.43708 
.42435 

.48557 
.4(5015 
.45329 

.43796 
.42315 

.40884 

.39501 
.38165 
.36875 

.43883 
.42196 
.40573 

.39012 
.37512 
.36069 

.34682 
.33348 
.32065 

.30679 
.37970 
.36335 

.34770 
.33273 
.31840 

.30469 
.29157 
.27902 

.35894 
.34185 
.32557 

.31007 
.29530 
.28124 

.26785 
.25509 
.24295 

.24151 
.22571 
.21095 

.19715 
.18425 
.17220 

.16093 
.15040 
.14056 

30 

31 
32 
33 

34 
35 
36 

37 
38 
39 

.55207 

.47674 

.41199 

.35628 

.30832 

.26700 

.23138 

.17411 

.13137 

.54125 
.5.3063 
.52023 

.51003 
.50003 
.4iK)22 

.48061 
.47119 
.46195 

.46511 
.45377 
.44270 

.43191 
.42137 
.41109 

.40107 
.39128 
.38174 

.39999 
.38834 
.37703 

.36604 
.35538 
.34503 

.33498 
.32523 
.31575 

.34423 
.33259 
.32134 

.31048 
.29i)98 
.28983 

.28003 
.27056 
.26141 

.29646 
.28506 
.27409 

.26355 
.25342 
.24367 

.23430 
.22529 
.21662 

.25550 
.24450 
.23397 

.22390 
.21425 
.20503 

.19620 
.18775 
.17967 

.22036 

.20987 
.19987 

.19035 
.18129 
.17266 

.16444 
.15661 
.14915 

.16425 
.15496 
.14619 

.13791 
.13011 
.12274 

.11580 
.10924 
.10306 

.12277 
.11474 
.10723 

.10022 
.09366 
.08754 

.08181 
.07(546 
.07146 

40 

41 
42 
43 

44 
45 
46 

47 
48 
49 

.45289 

.37243 

.36335 
.35448 
.34584 

.33740 
.32917 
.32115 

.31331 
.30567 
.29822 

.290i)4 

.30656 

.29763 
.28896 
.28054 

.27237 
.26444 
.25674 

.24926 
.24200 
.23495 

.25257 

.20829 

.17193 

.14205 

.09722 

.06678 

.44401 
.43530 

.42677 

.41840 
.41020 
.40215 

.39427 
.(«654 
.37896 

.24403 
.23578 
.22781 

.22010 
.21266 
.20547 

.19852 
.19181 
.18532 

.20028 
.19257 
.18517 

.17805 
.17120 
.16461 

.15828 
.15219 
.14634 

.16453 
.15744 
.15066 

.14417 
.13796 
.13202 

.12634 
.120i)0 
.11569 

.13528 
.12884 
.12270 

.11686 
.11130 
.10600 

.10095 
.0<M514 
.09156 

.09172 
.08653 
.08163 

.07701 
.07265 
.06854 

.06466 

.0(noo 

.05755 
.05429 

.06241 
.05833 
.05451 

.05095 
.04761 
.04450 

.04159 

.03887 
.03632 

50 

.37153 

.22811 

.17905 

.14071 

.11071 

.08720 

.03395 

INDEX 


[Numbers  refer  to  pages.l 


Abscissa  of  coordinates,  105. 

Accuracy  of  measurements,  four-fig., 
five-fig.,  etc.,  11. 

Acre,  Spanish,  U.  S.,  220. 

Acute  angle,  19. 

Addition  (subtraction) ,  66 ;  of  frac- 
tions, 73. 

Algebra,  first  use  of,  63;  review  of, 
63  ff. 

Algebraic  scale,  88. 

Alidade,  the,  on  a  transit,  220. 

Altitudes  of  a  triangle,  computation 
of,  186. 

Anchor  ring  or  torus,  197. 

Angle,  acute,  right,  obtuse,  19 ;  be- 
tween tangent  and  chord,  28;  be- 
tween two  chords,  188;  bisector 
of  an,  185;  central,  inscribed,  of  a 
circle,  27  ;  complementary  angles, 
ratios  for,  62 ;  construction  of  an, 
61 ;  definition  of  an,  19 ;  of  de- 
pression, of  elevation,  54,  219;  of 
incidence,  5 ;  interior,  of  a  regular 
polygon,  24 ;  polyhedral,  37 ;  of 
reflection,  5 ;  related  to  sides  of  a 
right  triangle,  46,  47  ;  sine,  cosine, 
tangent  of  an,  47 ;  straight,  19  ; 
trigonometric  ratios  of  an,  47 ;  a 
unit  of  measure  of  an,  19,  30 ;  ver- 
tex of  an,  19. 

Annuities,  151 ;  amount  of,  152 ; 
formulas  for,  164,  165 ;  present 
price  of,  155;  present  worth  of, 
154. 

Apothem,  23. 

Approximations,  accuracy  of,  11 ; 
in  measuring,  10,  122. 

Arc,  radius,  and  circumference  of  a 
circle,  28. 

Area,  charts,  99,  102  ;  of  a  circle,  28 ; 
of  an  ellipse,  191 ;  of  a  farm  sur- 
vey, 56  ff .,  222 ;  of  an  irregular 
shape,    24 ;     of    a    parallelogram, 


rectangle,  square,  trapezoid,  22; 
of  a  regular  polygon,  24 ;  of  a  sec- 
tor of  a  circle,  29 ;  of  a  segment, 
29,  189 ;   of  a  triangle,  20,  186. 

Arithmetic,  progression,  139 ;  aver- 
age, mean,  169. 

Arm,  moment,  235. 

Averages,  arithmetic,  geometric,  169 ; 
different  kinds  of,  171 ;  the  median, 
the  mode  as,  171,  176;  and  mix- 
tures, chapter  X,  167 ;  weighted, 
172. 

Axes  of  reference,  89. 

Balancing  a  survey,  59,  224. 
Bank  discount,  18. 
Bar  diagrams,  100. 
Barrel,  volume  of  a,  36. 
Base,  lines  of  a  survey,  31,  233;    of 
logarithms,  125  ;  in  percentage,  12. 
Bearing  of  a  line  or  course,  56. 
Binomial  theorem,  71. 
Bisector  of  an  angle  of  a  triangle,  185. 
Block  and  tackle,  241. 
Butter  fat  in  milk,  65,  167. 

Capitalization,  157. 

Cement,  unit  volume  of,  181. 

Center,  of  circumscribed,  inscribed 
circles  in  a  triangle,  185 ;  of  grav- 
ity, 185 ;   of  a  regular  polygon,  23. 

Centigrade,  thermometer,  compared 
with  Fahrenheit,  76,  109. 

Central  angle  of  a  circle,  27. 

Chain  used  by  surveyors,  218. 

Characteristic  of  a  logarithm,  126. 

Charts,  area,  99,  102. 

Checking  calculations  by  drawing  to 
scale,  6,  205 ;  by  the  law  of  sines, 
206  ff. 

Circles,  27;  circumscribed,  inscribed 
in  a  polygon,  27 ;  in  a  triangle, 
185 ;     great,    small    on    a   sphere, 


285 


286 


INDEX 


194;     secant,    tangent   to,    sector, 

segment  of,  27,  187. 
Circumference    of    a    circle,    28 ;     to 

draw  a  line  segment  equal  to  the, 

190. 
Clearing  an  equation  of  fractions,  75. 
Closed  survey,  57,  224. 
Common  logarithms  (base  ten),  125. 
Complementary  angles,  ratios  for,  62. 
Completing  the  square,  78. 
Components,    of    a    force,    9,    255 ; 

rectangular,  254  ;   of  the  resultant, 

251. 
Composition   of   forces,    of   motions, 

251. 
Compound  curves,  189. 
Compound  interest,    149 ;     annuities 

at,    151    ff.;    formulas,    164,    165; 

tables,  273. 
Computation,     10;      by    logarithms, 

124  ff. 
Concrete,  mixing  of,  180  ff. ;  Fuller's 

rule    for,     182 ;      proportions    for, 

for    various    purposes,    182 ;     unit 

volume  of,  181. 
Cone,  34 ;    frustum  of  a,  35 ;    slant 

height  of  a,  34 ;    surface  of  a,  35. 
Construction  of  angles,  51. 
Coordinate  paper,  90. 
Coordinates,  rectangular,  104. 
Corners  of  a  survey,  56. 
Correction  lines  or  standard  parallels, 

232. 
Cosine,  defined,  47;    law  of  cosines, 

204;   of  obtuse  angles,  112. 
Crane,  simple  jib,  258. 
Cube,    32;     cube    and    cube    roots, 

table  of,  275. 
Curve,  90 ;  compound,  189 ;  price,  98. 
Cylinder,  altitude  of,  33;    right,  34. 
Cylindrical  surface,  33. 

Dairymen's  parallelogram,  174. 

Dairy  problems,  167  ff . 

Data,  comparison  of,  90;  graphs  of, 
89. 

Datum  plane  in  leveling,  95. 

Decagon,  23. 

Degree,  19. 

Denominator  of  a  fraction,  72. 

Depreciation,  158 ;  formulas  for, 
164,165;  reducing  balance  method 
for,  160 ;   sinking  fund  method  for, 


162 ;     straight    line    method    for, 

159. 
Diagonal  scale,  draftsmen's,  45. 
Difference  of  latitudes,  of  longitudes, 

57,  222,  223. 
Directed  line  segment,  88. 
Direction  by  compass,  56. 
Discount,     bank,     18 ;      trade,     14 ; 

true,  17. 
Discounting  a  note,  18. 
Dividers,  hairspring,  2. 
Division  of  fractions,  73. 
Dot  diagrams,  103. 
Double  longitude  of  a  course,   222, 

223. 
Doubletree,    two-horse    evener,    267. 
Draft,  definition  of*  248;    down,  up 

a  grade,  249. 
Draftsmen's  diagonal  scale,  45. 
Drainage,  rules  for,  269. 
Drawing,  board,  1 ;   instruments  and 

materials  for,  1 ;   to  scale,  6. 

Easting  on  a  course,  57,  222. 

Ellipse,  area,  axes,  definition,  foci, 
perimeter,  191,  192. 

ElHpsoid,  196. 

Equation,  clearing  of  fractions,  75 ; 
definition,  63 ;  linear,  in  one 
unknown,  74-;  permissible  opera- 
tions, 64 ;  root,  63 ;  satisfying 
an  equation,  63 ;  simultaneous 
equations,  79  ;   solution,  63. 

Error,  small,  in  a  sum,  product, 
quotient,  85 ;  balancing  the  error, 
in  a  survey,  59,  224. 

Estimating  heights  and  distances, 
43,  44,  45. 

Exponents,  defined,  68 ;  fractional, 
80;  laws  of,  82;  negative,  81; 
table  of,  124. 

Exterior  angle,  184. 

Factor  of  safety  for  beams,  242. 

Factoring,  69. 

Fahrenheit  thermometer,  compared 
with  Centigrade,  76,  109. 

Farm,  loans,  151 ;  surveys,  56  ff.,  218. 

Fertilizers,  177  ff. ;  analysis  of 
materials  for,  271 ;  for  various 
crops.  Table  III,  Appendix. 

Figures,  significant,  11. 

Foot-pound  of  work,  defined,  245. 


INDEX 


287 


Force,  motion,  etc.,  235;  represented 
by  a  line  segment,  8. 

Forces,  graphic  representation,  reso- 
lution, etc.,  8,  251 ;  polygon  of, 
262;  resultant  of,  251,  252,  261, 
264  ;   triangle  of,  258. 

Fractional  exponents,  80. 

Fractions,  72  ;   complex,  74. 

Frustums  of  cones,  pyramids,  etc., 
33-35,  198. 

Geometric  progression,  142  ;    applied 

to  annuities,  151 ;   as  an  average  or 

mean,  169 ;   infinite,  146. 
Government    surveys,    general    plan 

of,  229. 
Grade    or    slope,    defined,    7,     114; 

draft  up  or  down  a,  249  ;   per  cent, 

7. 
Graphic    representation,    of    forces, 

251 ;    of  quantities,  88 ;    resolution 

of  forces  by,  8;    solutions  by,  1,  6. 
Graphs,  in  algebra,  104  ff. ;   of  data, 

89,  90 ;   of  equations,  106 ;  of  time, 

rate,  distance,  121. 

Hexagon,  23. 
Hexahedron,  30. 

Horizontal,  component,  9,  255;    pro- 
jection of  a  line  segment,  6. 
Horse-power,  definition  of,  246. 

Wcidence,  angle  of,  6. 

Inclined  plane,  forces  on  an,  9,  237. 

Indirect  measurement  and  propor- 
tion, 43,  44. 

Indorsement  of  a  note,  16. 

Infant  mortality,  table  of,  93. 

Inscribed  angle  in  a  circle,  27. 

Intercepts  on  the  axes,  110. 

Interest,  compound,  149 ;  exact,  16  ; 
formulas  for,  summary  of,  164 ; 
simple,  16. 

Interpolation,   129, 

Irregular  areas,  24. 

Jib  crane,  25$. 

Labor,  Spanish  land  unit,  220. 
Land,   computing  the  area  of,   222 ; 

locating,    by    metes    and    bounds, 

229;   surveying,  218  fif. 


Latitude-,    longitude-dififerences,    57, 

222,  223. 
Law,  of  cosines,  204;    of  levers.  111, 

236 ;   of  segments   in  the  solution 

of  triangles,  212  ;    of  sines,  203. 
League,  Spanish  land  unit,  220. 
Legal  milk,  167. 
Letters  and  figures,  set  of,  3. 
Level,  the,  95. 

Leveling,  described,  95 ;    rod,  218. 
Levers,  law  of.  111,  236. 
Linear    equation    in    one    unknown, 

74  ;   graph  of,  106. 
Line    segments,     3,    4 ;      comparing 

data   by,    99,    100;     directed,    88; 

projection  of,  6. 
Loading  beams,  ways  of,  243. 
Loans,  farm,  151. 
Locating  land  by  metes  and  bounds, 

229. 
Logarithmic  scale,  133. 
Logarithms,  characteristic,  mantissa 

of,  126 ;   common,  to  base  ten,  125 ; 

computation     by,     124    ff.,    204; 

defined,     125 ;      interpolation     in, 

129;    table  of,  explained,   128. 
Longitude-,    latitude-differences,    57, 

222,  223. 

Mannheim  slide  rule,  134. 
Mantissa  of  a  logarithm,  126. 
Mean,    arithmetic,    geometric,    169 ; 

proportional,  170;    weighted,   172. 
Measurements,    are    approximations, 

10,    122 ;    errors  in,   84 ;    indirect, 

43 ;     units   of,    in   surveying,    218, 

220. 
Measuring  angles  with  a  protractor, 

2. 
Mechanical  advantage,  defined,  240. 
Median,  176;   of  a  triangle,  185. 
Mensuration,  elementary,  18. 
Meridians,     base    lines,     etc.,     230; 

principal,  233. 
Metes    and    bounds,    locating    land 

by,  229. 
Milk,   can,   finding  volume  of,    196 ; 

standard,  legal,  167. 
Mixtures,     and     averages,     chapter 

X,  167  ;   standardizing  method  for, 

174. 
Mode,  176. 
Moment,  arm,  235 ;    of  a  force,  235. 


288 


INDEX 


Mortality  table,  Table  VI,  Appendix. 

Motion,  force,  235 ;  composition  of 
motions,  of  forces,  251. 

Multiplication,  of  fractions,  73  ;  type- 
forms  in,  68 ;  by  use  of  slide  rule, 
136. 

Negative  exponent,  81. 
Northing,  57,  222,  224. 
Number  scale,  88. 
Numerator  of  a  fraction,  72. 

Oblique  triangles,  184 ;    construction 

of,  200 ;   solution  of,  202. 
Obtuse     angle,     19 ;      sine,     cosine, 

tangent  of  an,  112. 
Octagon,  23. 
Octahedron,  30. 
Operations  permitted  on  an  equation, 

64. 
Ordinate  of  coordinates,  105. 
Origin  of  coordinates,  105. 
Orthocenter,  185. 

Paper,  coordinate,  90 ;    profile,  96. 

Paraboloid,  196. 

Parallel,  lines,  slopes  of,  119;  forces, 
resultant  of,  264. 

Parallelepiped,  32. 

Parallelogram,  22  ;  dairymen's,  174  ; 
of  forces,  251 ;  method  for  stand- 
ardizing mixtures.  174. 

Parallels,  standard,  or  correction 
lines,  232. 

Partial  payments,  U.  S.  Rule  for,  16. 

Pentagon,  23. 

Percentage,  12. 

Per  cent,  grade,  7;   rate  in.  12. 

Perimeter,  of  an  ellipse,  192 ;  of  a 
polygon,  23. 

Perpendicular  lines,  120. 

Perpetuities,  157 ;  present  value  of, 
157. 

Perspective  of  similar  shapes,  37. 

TT  (pi),  28. 

Pitch,  of  a  roof,  7 ;    of  a  screw,  23. 

Plat  of  a  survey,  218. 

Polygon,  apothem  of,  23 ;  center  of 
regular,  23  ;  definition  of,  23 ;  of 
forces,  262 ;  inscribed  in  a  circle, 
27  ;    perimeter  of,  23  ;    regular,  23. 

Polyhedral  angle,  37. 

Polyhedrons,  30. 


Present  worth,  17 ;  of  an  annuity,  154. 

Price,  of  an  annuity,  155  ;   curves,  98. 

Principal,  in  simple  interest,  16 ; 
meridians,  231,  233. 

Prismoid  formula,  197. 

Prisms,  31 ;  regular,  right,  31 ; 
truncated,  193. 

Proceeds  of  a  note,  18. 

Profile,  of  a  course  in  leveling,  96 ; 
paper,  96. 

Progression,  arithmetic,  139 ;  geo- 
metric, 142  ;  infinite  geometric,  146. 

Projection,  of  a  line  segment,  6 ; 
horizontal,  6 ;  vertical,  7 ;  of  the 
sides  of  a  triangle,  186. 

Proportion,  defined,  43. 

Proportional,  mean,  170. 

Protractor,  2. 

Pulleys,  systems  of,  240. 

Pyramid,  32  ;  altitude,  slant  height, 
vertex  of  a,  32 ;    frustum  of  a,  33. 

Pythagoras  Theorem,  the,  20;  gen- 
eralized for  oblique  triangles,  185. 

Quadrants,    the   plane   divided   into, 

105. 
Quadratic  equations,  solution  of,  77. 
Quadrilaterals,  22. 

Radian,  30. 

Radicals,  combining,  84 ;  reduction 
of,  83  ;   similar,  83. 

Radius,  28. 

Range,  poles,  218 ;   of  townships,  230. 

Rate,  per  cent,  12  ;  in  simple  interest, 
16 ;   of  doing  work,  246. 

Ratio,  defined,  43 ;  of  terms  in  a 
geometric  progression,   142. 

Ratios,  for  complementary  angles, 
62;  for  obtuse  angles,  112;  trig- 
onometric, 47. 

Rectangle,  22. 

Rectangular,  components  of  a  force, 
254 ;   coordinates,  104. 

Reducing  balance  method  for  depre- 
ciation, 160,  161. 

Reflection,  angle  of,  6. 

Regular  polygon,  23 ;   area  of,  24. 

Resolution  of  forces,  254  ;   graphic,  8. 

Resultant  of  forces,  251,  252,  261; 
of  parallel  forces,  264. 

Revolution,  a  complete,  19. 

Right  angle,  19. 


INDEX 


289 


Right  section  of  a  prism,  31. 

Right  triangle,  altitude,  sides  of, 
184 ;  the  Pythagoras  Theorem  for 
the,  20,  184;  relations  of  sides  and 
angles  in  a,  46,  47 ;  solution  of 
the,  52 ;  trigonometric  ratios  for 
the,  47. 

Rise,  of  a  slope  or  grade,  7. 

Root  of  an  equation,  63. 

Run,  of  a  slope  or  grade,  7. 

Scale,  algebraic,  number,  88;  drafts- 
men's diagonal,  45;  drawing  to, 
6 ;  linear,  2 ;  logarithmic,  133 ; 
on  slide  rule,  135. 

Screw,  pitch  of  a,  general  principle 
of,  237. 

Secant  line  for  a  circle,  27 ;  and 
tangent  from  an  external  point, 
188. 

Section  of  land,  220,  231. 

Sector  of  a  circle,  27. 

Segment,  of  a  circle,  27,  189 ;  law  of 
segments  in  the  solution  of  tri- 
angles, 212;  line,  3,  4,  99;  of  a 
sphere,  195. 

Significant  figures,  11. 

Silos,  41. 

Similar,  figures,  solids,  surfaces,  37 ; 
terms,  66. 

Simple  interest,  16. 

Simultaneous  equations,  79;  solu- 
tion of,  110. 

Sine,  defined,  47 ;  law  of  sines,  203 ; 
of  obtuse  angles,  112. 

Sinking  fund,  156 ;  method  of  depre- 
ciation, 162. 

Slide  rule,  description  and  history, 
134 ;  Mannheim,  the,  134  ;  paper, 
to  be  cut  from  the  last  fly  leaf,  135 ; 
runner,  slide,  stock  of  a,  134 ; 
scales  of  a,  135. 

Slope,  measured  by  rise  divided  by 
run,  7,  114;  positive,  negative, 
115;   of  a  straight  line,  114. 

Solids,  surfaces  and  volumes  of,  35, 
36. 

Solutions,  checking  of,  205 ;  of  an 
equation,  63;  graphic,  1,  6;  of 
oblique  triangles,  202  ;  of  quadratic 
equations,  77;  of  right  triangles, 
52 ;  of  simultaneous  equations, 
110 ;  summary  of,  of  triangles,  216. 


Southing  of  a  course,  57,  222,  224. 

Span  of  a  roof,  7. 

Sphere,  35,  194,  195. 

Spherical  segment,  volume  of,  195. 

Square,  22. 

Squared  paper,  51,  90. 

Squares,  square  roots,  on  slide  rule, 

138 :  table  of,  Table  VII,  Appendix. 
Straight  angle,  19. 
Straight  line  method  of  depreciation, 

159. 
Straight    lines,     equations    of,     116, 

117;    graphs  of,   106  ff. ;    parallel, 

119;     perpendicular,    120;     slopes 

of,  114. 
Strength  of  beams,  of  materials,  242. 
Surfaces  and  volumes  of  soHds,  35,  36. 
Surveys,  balancing,  59,  224 ;    closed, 

57,  224 ;    computing  the  area  of, 

58,  222 ;    farm,   56 ;    government, 
general  plan  of,  230;    land,  218. 

Tables,  of  logarithms,  128  and  Ap- 
pendix ;  of  trigonometric  ratios, 
49  and  Appendix ;  reading  of,  for 
obtuse  angles,  113. 

Tangent,  of  an  angle,  defined,  47, 
112;  to  a  circle,  27;  and  secant 
from  an  external  point,  188. 

Tetrahedron,  300. 

Theorem,  binomial,  71. 

Time,  rate,  distance  graphs,  121. 

Torus,  anchor  ring,  197. 

Townships,  230,  232. 

Trade  discounts,  14. 

Transit,  the,  for  measuring  angles,  219. 

Trapezoid,  22. 

Triangle,  altitudes,  medians,  sides, 
etc.,  184,  185;  area  of  a,  20,  186; 
of  forces,  258 ;  equilateral,  isosceles, 
right,  etc.,  20  ;  triangles  for  graphic 
work,  2. 

Trigonometric  ratios,  for  comple- 
mentary angles,  62  ;  defined,  47 ; 
for  obtuse  angles,  112,  113;  tables 
of,  49,  275. 

True  discount,  17. 

Unit  of  volume,  32. 

Units    and    equivalents    in    weights 

and  measures,  271. 
Units  of  measurements  in  surveying, 

218,  220. 


290 


INDEX 


Vara,  Spanish  land  unit,  220. 

Vertex  of  an  angle,  19. 

Vertical,  component,  9,  255;  projec- 
tion of  a  line  segment,  7- 

Visibility  from  elevations  above  the 
earth,  rule  for,  93,  188. 

Volume,  and  surfaces  of  solids,  35,  36 ; 
of  a  truncated  prism,  193  ;unit  of,  32. 


Watt,    James,    experiments 
dray  horses,  246. 


of. 


Weighted  arithmetic  mean,  172. 
Weights,    and    measures,    tables    of, 

271 ;   of  produce,  269. 
Westing  on  a  course,  57,  222. 
Wheel  and  axle,  law  of,  236. 
Work,   foot-pound  of,   245 ;    general 

principle  of,  237,  245,  248. 
Working  day  for  a  horse,  246. 

Zero  exponent,  81. 
Zone  of  a  sphere,  195. 


SLIDE  RULE 


!  I  P 

i 


y%. 


I 


(J)  (^)  (^) 


A  reasonably  accurate  slide  rule 
may  be  made  by  the  student,  for 
temporary  practice,  as  follows. 
Take  three  strips  of  heavy,  stiff 
cardboard  1''.3  wide  by  6"  long; 
these  are  shown  in  cross-section  in 
(1),  (2),  (3)  above.  On  (3) 
paste  or  glue  the  adjoining  cut 
of  the  slide  rule.  Then  cut  strips 
(2)  and  (3)  accurately  along  the 
lines  marked.  Paste  or  glue  the 
pieces  together  as  shown  in  (4) 
and  (5).  Then  (5)  forms  the 
slide  of  the  slide  rule,  and  it  will 
fit  in  the  groove  in  (4)  if  the  work 
has  been  carefully  done.  Trim 
off  the  ends  as  shown  in  the  large 
out. 


( 

< 

ca     o 

0> 
CO 

- 

E 

= 

00 

0>- 

=  E 

-a 

rrt 

— 

= 

tr- 

t- 

EE 

— 

— 

=: 

- 

CD 

EE 

- 

i 

1 

io 

lO 

::  = 

^ 

CO  ■ 

-  — 

-o 

1 

1 

iG- 

=  = 

-o 

CO 

— 

— 

1 

1 

= 

E 

-- 

= 

= 

= 

^ 

^: 

-rp 

— 

— 

— 

— 



_ 

_ 

— 

• 

eo- 

- 

- 

tH 

= 

— 

zz. 

ZI 

- 

E 

E 

-CO 

00 

= 

E 

<0 

- 1 

t- 

- 

- 

= 

E 

<0 

1 

E 

^ 

Is 

1 

= 

i 

E 

« 

E 

E 

5 

E 

—i 

— 

-^ 

— 

^ 

— 

— 

— 

- 

- 

- 

- 

^ 

< 

CQ        O 

Q 

J 

THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED   FOR   FAILURE  TO   RETURN 
THIS    BOOK  ON   THE   DATE   DUE.    THE   PENALTY 
WILL  INCREASE  TO  50  CENTS  ON  THE  FOURTH 
DAY    AND     TO     $1.00     ON     THE    SEVENTH     DAY 
OVERDUE. 

y'^f^M. 

^T^ 

LD  21-100m-7.'39(402s) 


MM 


VM 
li 

1,1,1  'nihi,!:  ,|i 


'lIlilH 


:1 
v>l'ilii 


iHlj! 


lihi.'i.lii''-'* 


iMil   'i   I. 

;ti'!i|!!i||i!n 


"i  1 


'ffl 

.11 


lipi! 
'„!iiii|li!i|! 


n 


I  !P  P    P 

ii 

j|t|i||Hm 


,n.i 
'i 


iiliriiijij 


t  1  i-s 


iM|il!!!lir 


i 
m 


! 


i  ill 

lliiliilliliiilil!! 


! 


ipliiiiii! 


mm 
li  pffli 
mwm 

ft  mmu 
1  pllj 

mm 

iliililip 


Wmm 


111 


li 


'Villllipll 


^mmi 


1  _      Mil 


iiii'iiiiiii  iiili 


tl 


'I 


wmm 


II 


